Chapter 12
Static Equilibrium and Elasticity
Conceptual Problems
1
(a)
[SSM] True or false:
F
∑ i = 0 is sufficient for static equilibrium to exist.
•
i
(b)
∑ F = 0 is necessary for static equilibrium to exist.
i
i
(c)
(d)
In static equilibrium, the net torque about any point is zero.
An object in equilibrium cannot be moving.
(a) False. The conditions
∑ F = 0 and ∑τ
= 0 must be satisfied.
(b) True. The necessary and sufficient conditions for static equilibrium are
∑ F = 0 and ∑τ = 0 .
(c) True. The conditions
∑ F = 0 and ∑τ
= 0 must be satisfied.
(d) False. An object can be moving with constant speed (translational or rotational)
when the conditions ∑ F = 0 and ∑τ = 0 are satisfied.
2
•
True or false:
(a) The center of gravity is always at the geometric center of a body.
(b) The center of gravity must be located inside an object.
(c) The center of gravity of a baton is located between the two ends.
(d) The torque produced by the force of gravity about the center of gravity is always
zero.
(a) False. The location of the center of gravity depends on how an object’s mass is
distributed.
(b) False. An example of an object for which the center of gravity is outside the
object is a donut.
(c) True. The structure of a baton and the definition of the center of gravity
guarantee that the center of gravity of a baton is located between the two ends.
(d) True. Because the force of gravity acting on an object acts through the center of
gravity of the object, its lever (or moment) arm is always zero.
1185
1186 Chapter 12
3
•
The horizontal bar in Figure 12-27 will remain horizontal if (a) L1 = L2
and R1 = R2, (b) M1R1 = M2R2, (c) M2R1 = R2M1, (d) L1M1 = L2M2,
(e) R1L1 = R2L2.
Determine the Concept The condition that the bar is in rotational equilibrium is
that the net torque acting on it equal zero; i.e., R1M1 − R2M2 = 0. (b) is correct.
4
•
Sit in a chair with your back straight. Now try to stand up without
leaning forward. Explain why you cannot do it.
Determine the Concept You cannot stand up because, if you are to stand up, your
body’s center of gravity must be above your feet.
5
•
You have a job digging holes for posts to support signs for a Louisiana
restaurant (called Mosca’s). Explain why the higher above the ground a sign is
mounted, the farther the posts should extend into the ground.
Determine the Concept Flat signs of any kind experience substantial forces
when the wind blows against them – the larger the surface area, the larger the
force. In order to be stable, the posts which support such signs must be buried
deeply enough so that the ground can exert sufficient force against the posts to
keep the sign in equilibrium under the strongest winds. The pivot point around
which the sign might rotate is at ground level – thus the more moment arm
available below ground level, the more torque may be generated by the force of
the ground on the posts. Thus the larger the surface area of the billboard, the
greater will be the force applied above the surface, and hence the torque applied
to the posts will be greater. As surface area increases, the preferred depth of the
posts increases as well so that with the increased moment arm, the ground can
exert more torque to balance the torque due to the wind.
6
•
A father (mass M) and his son, (mass m) begin walking out towards
opposite ends of a balanced see-saw. As they walk, the see-saw stays exactly
horizontal. What can be said about the relationship between the father’s speed V
and the son’s speed v?
Determine the Concept The question is about a situation in which an object is in
static equilibrium. Both the father and son are walking outward from the center of
the see-saw, which always remains in equilibrium. In order for this to happen, at
any time, the net torque about any point (let’s say, the pivot point at the center of
the see-saw) must be zero. We can denote the father’s position as X, and the son’s
position as x, and choose the origin of coordinates to be at the pivot point. At each
moment, the see-saw exerts normal forces on the son and his father equal to their
respective weights, mg and Mg. By Newton’s third law, the father exerts a
downward force equal in magnitude to the normal force, and the son exerts a
downward force equal in magnitude to the normal force acting on him.
Static Equilibrium and Elasticity 1187
Apply
∑τ
pivot point
= 0 to the see-saw
MgX − mgx = 0
(1)
(assume that the father walks to the
left and that counterclockwise
torques are positive):
Express the distance both the father
and his son walk as a function of
time:
Substitute for X and x in equation (1)
to obtain:
X = VΔt and x = vΔt
MgVΔt − mgvΔt = 0 ⇒ V =
m
v
M
Remarks: The father’s speed is less than the son’s speed by a factor of m/M.
7
•
Travel mugs that people might set on the dashboards of their cars are
often made with broad bases and relatively narrow mouths. Why would travel
mugs be designed with this shape, rather than have the roughly cylindrical shape
that mugs normally have?
Determine the Concept The main reason this is done is to lower the center of
gravity of the mug as a whole. For a given volume, it is possible to make a mug
with gently sloping sides that has a significantly lower center of gravity than the
traditional cylinder. This is important, because as the center of gravity of an object
gets lower (and as its base broadens) the object is harder to tip. When cars are
traveling at constant velocity, the design of the mug is not important – but when
cars are stopping and going – accelerating and decelerating – the higher center of
gravity of the usual design makes it much more prone to tipping.
8
••
The sailors in the photo are using a technique called ″hiking out″. What
purpose does positioning themselves in this way serve? If the wind were stronger,
what would they need to do in order to keep their craft stable?
Determine the Concept Dynamically the boats are in equilibrium along their line
of motion, but in the plane of their sail and the sailors, they are in static equilibrium.
The torque on the boat, applied by the wind acting on the sail, has a tendency to tip
the boat. The rudder counteracts that tendency to some degree, but in particularly
strong winds, when the boat is sailing at particular angles with respect to the wind,
the sailors need to ″hike out″ to apply some torque (due to the gravitational force of
the Earth on the sailors) by leaning outward on the beam of the boat. If the wind
strengthens, they need to extend their bodies further over the side and may need to
get into a contraption called a ″trapeze″ that enables the sailor to have his or her
entire body outside the boat.
1188 Chapter 12
9
••
[SSM] An aluminum wire and a steel wire of the same length L and
diameter D are joined end-to-end to form a wire of length 2L. One end of the wire is
then fastened to the ceiling and an object of mass M is attached to the other end.
Neglecting the mass of the wires, which of the following statements is true? (a) The
aluminum portion will stretch by the same amount as the steel portion. (b) The
tensions in the aluminum portion and the steel portion are equal. (c) The tension in
the aluminum portion is greater than that in the steel portion. (d) None of the above
Determine the Concept We know that equal lengths of aluminum and steel wire
of the same diameter will stretch different amounts when subjected to the same
tension. Also, because we are neglecting the mass of the wires, the tension in
them is independent of which one is closer to the roof and depends only on Mg.
(b) is correct.
Estimation and Approximation
10 ••
A large crate weighing 4500 N rests on four 12-cm-high blocks on a
horizontal surface (Figure 12-28). The crate is 2.0 m long, 1.2 m high and 1.2 m
deep. You are asked to lift one end of the crate using a long steel pry bar. The
fulcrum on the pry bar is 10 cm from the end that lifts the crate. Estimate the length
of the bar you will need to lift the end of the crate.
Picture the Problem The diagram to
the right shows the forces acting on
the crate as it is being lifted at its left
end. Note that when the crowbar lifts
the crate, only half the weight of the
crate is supported by the bar. Choose
the coordinate system shown and let
the subscript ″pb″ refer to the pry bar.
The diagram below shows the forces
acting on the pry bar as it is being used
to lift the end of the crate.
y
r
W
Frpb
Fpb
Fn
W
w
B
y
r
F'
l
F
B
x
A
L−l
Fpb
x
Static Equilibrium and Elasticity 1189
Assume that the maximum force F you can apply is 500 N (about 110 lb). Let be
the distance between the points of contact of the steel bar with the floor and the
crate, and let L be the total length of the bar. Lacking information regarding the
bend in pry bar at the fulcrum, we’ll assume that it is small enough to be negligible.
We can apply the condition for rotational equilibrium to the pry bar and a condition
for translational equilibrium to the crate when its left end is on the verge of lifting.
Apply
∑F
Apply
∑τ
y
= 0 to the crate:
= 0 to the crate about an
Fpb − W + Fn = 0
(1)
wFn − 12 wW = 0 ⇒ Fn = 12 W
axis through point B and perpendicular
to the plane of the page to obtain:
as noted in Picture the Problem.
Solve equation (1) for Fpb and
substitute for Fn to obtain:
Fpb = W − 12 W = 12 W
Apply
∑τ
= 0 to the pry bar about
an axis through point A and
perpendicular to the plane of the
page to obtain:
F (L −
)−
⎛ F ⎞
Fpb = 0 ⇒ L = ⎜⎜1 + pb ⎟⎟
F ⎠
⎝
Substitute for Fpb to obtain:
⎛ W ⎞
L = ⎜1 +
⎟
⎝ 2F ⎠
Substitute numerical values and
evaluate L:
⎛
4500 N ⎞
⎟⎟ = 55 cm
L = (0.10 m )⎜⎜1 +
⎝ 2(500 N ) ⎠
11 ••
[SSM] Consider an atomic model for Young’s modulus. Assume that
a large number of atoms are arranged in a cubic array, with each atom at a corner of
a cube and each atom at a distance a from its six nearest neighbors. Imagine that
each atom is attached to its 6 nearest neighbors by little springs each with force
constant k. (a) Show that this material, if stretched, will have a Young’s modulus
Y = k/a. (b) Using Table 12-1 and assuming that a ≈ 1.0 nm, estimate a typical value
for the ″atomic force constant″ k in a metal.
Picture the Problem We can derive this expression by imagining that we pull on
an area A of the given material, expressing the force each spring will experience,
finding the fractional change in length of the springs, and substituting in the
definition of Young’s modulus.
1190 Chapter 12
(a) The definition of Young’s
modulus is:
Y=
Express the elongation ΔL of each
spring:
F A
ΔL L
(1)
ΔL =
Fs
k
(2)
The force Fs each spring will
experience as a result of a force F
acting on the area A is:
Fs =
F
N
Express the number of springs N
in the area A:
N=
A
a2
Fs =
Fa 2
A
Substitute Fs in equation (2) to
obtain, for the extension of one
spring:
ΔL =
Fa 2
kA
Assuming that the springs
extend/compress linearly, the
fractional extension of the springs is:
ΔLtot ΔL 1 Fa 2 Fa
=
=
=
L
a
a kA
kA
Substitute in equation (1) and
simplify to obtain:
F
k
Y= A =
Fa
a
kA
(b) From our result in Part (a):
k = Ya
From Table 12-1:
Y = 200GN/m 2 = 2.00 × 1011 N/m 2
Substitute numerical values and
evaluate k:
k = 2.00 ×1011 N/m 2 1.0 ×10 −9 m
Substituting for N yields:
(
)(
)
= 2.0 N/cm
12 ••
By considering the torques about the centers of the ball joints in your
shoulders, estimate the force your deltoid muscles (those muscles on top of your
shoulder) must exert on your upper arm, in order to keep your arm held out and
extended at shoulder level. Then, estimate the force they must exert when you hold
a 10-lb weight out to the side at arm’s length.
Picture the Problem A model of your arm is shown in the pictorial representation.
Your shoulder joint is at point P and the force the deltoid muscles exert on your
Static Equilibrium and Elasticity 1191
extended arm Fdeltoid is shown acting at an angle θ with the horizontal. The weight
of your arm is the gravitational force Fg = m g exerted by Earth through the center
of gravity of your arm. We can use the condition for rotational equilibrium to
estimate the forces exerted by your deltoid muscles. Note that, because its moment
arm is zero, the torque due to Fshoulder about an axis through point P and
perpendicular to the page is zero.
r
Fdeltoid
P
r
Fshoulder
Apply
∑τ
P
L
θ
r
= 0 to your extended
r
r
Fg = m g
Fdeltoid sin θ − 12 mgL = 0
(1)
arm:
Solving for Fdeltoid yields:
Assuming that ≈ 20 cm, L ≈ 60 cm,
mg ≈ 10 lb, and θ ≈ 10°, substitute
numerical values and evaluate
Fdeltoid:
If you hold a 10-lb weight at the end
of your arm, equation (1) becomes:
Solving for Fdeltoid yields:
Substitute numerical values and
evaluate F ′deltoid:
Fdeltoid =
mgL
2 sin θ
Fdeltoid =
(10 lb)(60 cm) ≈
2(20 cm )sin 10°
86 lb
F'deltoid sin θ − 12 mgL − m'gL = 0
where m′ is the mass of the 10-lb
weight.
F' deltoid =
mgL + 2m'gL
2 sin θ
Fdeltoid =
(10 lb)(60 cm) + 2(10 lb)(60 cm)
2(20 cm )sin 10°
≈ 260 lb
Conditions for Equilibrium
13 •
Your crutch is pressed against the sidewalk with a force Fc along its
own direction, as shown in Figure 12-29. This force is balanced by the normal
1192 Chapter 12
force Fn and a frictional force fs . (a) Show that when the force of friction is at its
maximum value, the coefficient of friction is related to the angle θ by μs = tan θ.
(b) Explain how this result applies to the forces on your foot when you are not using
a crutch. (c) Why is it advantageous to take short steps when walking on slippery
surfaces?
Picture the Problem Choose a coordinate system in which upward is the positive y
direction and to the right is the positive x direction and use the conditions for
translational equilibrium.
(a) Apply
∑ F = 0 to the forces
acting on the tip of the crutch:
∑F
x
= − f s + Fc sin θ = 0
and
∑ Fy = Fn − Fc cosθ = 0
Solve equation (2) for Fn and
assuming that fs = fs,max, obtain:
f s = f s,max = μs Fn = μs Fc cos θ
Substitute in equation (1) and solve
for µs:
μs = tan θ
(1)
(2)
(b) Taking long strides requires a large coefficient of static friction because θ is
large for long strides.
(c) If μs is small (the surface is slippery), θ must be small to avoid slipping.
14 ••
A thin rod of mass M is suspended horizontally by two vertical wires.
One wire is at the left end of the rod, and the other wire is 2/3 of the way from the
left end. (a) Determine the tension in each wire. (b) An object is now hung by a
string attached to the right end of the rod. When this happens, it is noticed that the
wire remains horizontal but the tension in the wire on the left vanishes. Determine
the mass of the object.
Picture the Problem The pictorial representation shows the thin rod with the forces
described in Part (a) acting on it. We can apply ∑τ 0 = 0 to the rod to find the
forces TL and TR. The simplest way to determine the mass m of the object
suspended from the rod in (b) is to apply the condition for rotational equilibrium a
second time, but this time with respect to an axis perpendicular to the page and
through the point at which TR acts.
Static Equilibrium and Elasticity 1193
r
TL
1
2
r
TR
L
0
2
3
L
L
r
Mg
(a) Apply
Apply
∑τ
0
∑F
vertical
= 0 to the rod:
= 0 to the rod:
2
3
LTR − 12 LMg = 0 ⇒ TR =
3
4
Mg
TL − Mg + TR = 0
Substitute for TR to obtain:
TL − Mg + 34 Mg = 0 ⇒ TL =
(b) With an object of mass m
suspended from the right end of the
rod and TL = 0, applying ∑τ = 0
( 23 − 12 )LMg − (1 − 23 )Lmg = 0
1
4
Mg
about an axis perpendicular to the
page and through the point at which
TR acts yields:
Solving for m yields:
m=
1
2
M
The Center of Gravity
15 •
An automobile has 58 percent of its weight on the front wheels. The
front and back wheels on each side are separated by 2.0 m. Where is the center of
gravity located?
Picture the Problem Let the weight of the automobile be w. Choose a coordinate
system in which the origin is at the point of contact of the front wheels with the
ground and the positive x axis includes the point of contact of the rear wheels with
the ground. Apply the definition of the center of gravity to find its location.
Use the definition of the center of
gravity to obtain:
xcgW = ∑ wi xi
Because W = w:
xcg w = (0.84 m ) w ⇒ xcg = 84 cm
i
= 0.58w (0) + 0.42 w (2.0 m )
= (0.84 m ) w
1194 Chapter 12
Static Equilibrium
16 •
Figure 12-30 shows a lever of negligible mass with a vertical force Fapp
being applied to lift a load F. The mechanical advantage of the lever is defined as
M = F Fapp, min , where Fapp, min is the smallest force necessary to lift the load F.
Show that for this simple lever system, M = x/X, where x is the moment arm
(distance to the pivot) for the applied force and X is the moment arm for the load.
Picture the Problem We can use the given definition of the mechanical advantage
of a lever and the condition for rotational equilibrium to show that M = x/X.
Express the definition of
mechanical advantage for a lever:
M=
Apply the condition for rotational
equilibrium to the lever:
∑τ
Solve for the ratio of F to Fapp, min
F
(1)
Fapp , min
fulcrum
F
to obtain:
Fapp, min
Substitute for F Fapp, min in equation (1)
to obtain:
M =
=
= xFapp, min − XF = 0
x
X
x
X
17 •
[SSM]
Figure 12-31 shows a 25-foot sailboat. The mast is a uniform
120-kg pole that is supported on the deck and held fore and aft by wires as shown.
The tension in the forestay (wire leading to the bow) is 1000 N. Determine the
tension in the backstay (wire leading aft) and the normal force that the deck exerts
on the mast. (Assume that the frictional force the deck exerts on the mast to be
negligible.)
Picture the Problem The force diagram
shows the forces acting on the mast. Let
the origin of the coordinate system be at
the foot of the mast with the +x direction
to the right and the +y direction upward.
Because the mast is in equilibrium, we
can apply the conditions for translational
and rotational equilibrium to find the
tension in the backstay, TB, and the
normal force, FD, that the deck exerts on
the mast.
y
TF
TB
θF
45°
mg
FD
P
x
Static Equilibrium and Elasticity 1195
Apply
∑τ
= 0 to the mast about
an axis through point P:
Solve for TB to obtain:
(4.88 m )(1000 N )sin θ F
− (4.88 m )TB sin 45.0° = 0
TB =
(1000 N )sin θ F
(1)
sin 45.0°
Find θ F , the angle of the forestay
with the vertical:
θ F = tan −1 ⎜⎜
Substitute numerical values in
equation (1) and evaluate TB:
TB =
⎛ 2.74 m ⎞
⎟⎟ = 29.3°
⎝ 4.88 m ⎠
(1000 N )sin 29.3° =
692 N
sin 45.0°
Apply the condition for translational equilibrium in the y direction to the mast:
∑F
y
= FD − TF cosθ F − TB cos 45° − mg = 0
Solving for FD yields:
FD = TF cosθ F + TB cos 45° + mg
Substitute numerical values and evaluate FD:
(
)
FD = (1000 N )cos 29.3° + (692 N )cos 45° + (120 kg ) 9.81 m/s 2 = 2.54 kN
18 ••
A uniform 10.0-m beam of mass 300 kg extends over a ledge as in
Figure 12-32. The beam is not attached, but simply rests on the surface. A 60.0-kg
student intends to position the beam so that he can walk to the end of it. What is the
maximum distance the beam can extend past end of the ledge and still allow him to
perform this feat?
Picture the Problem The diagram
shows Mg, the weight of the beam,
mg, the weight of the student, and the
force the ledge exerts F , acting on the
beam. Because the beam is in
equilibrium, we can apply the condition
for rotational equilibrium to the beam to
find the location of the pivot point P that
will allow the student to walk to the end
of the beam.
Apply
∑τ
= 0 about an axis
through the pivot point P:
r
F
x
5.0 m
P
r
Mg
Mg (5.0 m − x ) − mgx = 0
r
mg
1196 Chapter 12
5.0 M
M +m
Solving for x yields:
x=
Substitute numerical values and
evaluate x:
x=
(5.0 m )(300 kg )
300 kg + 60.0 kg
= 4.2 m
19 ••
[SSM] A gravity board is a convenient and quick way to determine
the location of the center of gravity of a person. It consists of a horizontal board
supported by a fulcrum at one end and a scale at the other end. To demonstrate this
in class, your physics professor calls on you to lie horizontally on the board with the
top of your head directly above the fulcrum point as shown in Figure 12-33. The
scale is 2.00 m from the fulcrum. In preparation for this experiment, you had
accurately weighed yourself and determined your mass to be 70.0 kg. When you are
at rest on the gravity board, the scale advances 250 N beyond its reading when the
board is there by itself. Use this data to determine the location of your center of
gravity relative to your feet.
Picture the Problem The diagram
shows w , the weight of the student,
the pivot, and Fs , the force exerted
by the scale, acting on the student.
Because
the
student
is
in
equilibrium, we can apply the
condition for rotational equilibrium
to the student to find the location of
his center of gravity.
Apply
∑τ
= 0 about an axis
r
FS
r
FP
FP , the force exerted by the board at
2.00 m
P
x
r r
w = mg
Fs (2.00 m ) − wx = 0
through the pivot point P:
Solving for x yields:
Substitute numerical values and
evaluate x:
x=
x=
(2.00 m )Fs
w
(2.00 m )(250 N ) =
(70.0 kg )(9.81m/s2 )
0.728 m
20 ••
A stationary 3.0-m board of mass 5.0 kg is hinged at one end. A force
F is applied vertically at the other end, and the board makes at 30° angle with the
horizontal. A 60-kg block rests on the board 80 cm from the hinge as shown in
Figure 12-34. (a) Find the magnitude of the force F . (b) Find the force exerted by
Static Equilibrium and Elasticity 1197
the hinge. (c) Find the magnitude of the force F , as well as the force exerted by the
hinge, if F is exerted, instead, at right angles to the board.
Picture the Problem The diagram
shows mg, the weight of the board,
r
F
Fhinge , the force exerted by the hinge,
Mg, the weight of the block, and F ,
the force acting vertically at the right
end of the board. Because the board
is in equilibrium, we can apply the
condition for rotational equilibrium
to it to find the magnitude of F .
(a) Apply
0m
1.5
y
r
Fhinge
0
0.8
30°
P
0m
m
0.7
r
Mg
r
mg
x
∑τ = 0 about an axis through the hinge to obtain:
F [(3.0 m ) cos 30°] − mg [(1.50 m ) cos 30°] − Mg[(0.80 m ) cos 30°] = 0
Solving for F yields:
F=
m (1.50 m ) + M (0.80 m )
g
3.0 m
Substitute numerical values and evaluate F:
Apply
F=
(5.0 kg )(1.50 m ) + (60 kg )(0.80 m ) (9.81 m/s 2 ) = 181 N =
∑F
y
0.81 kN
3.0 m
= 0 to the board to
Fhinge − Mg − mg + F = 0
obtain:
Solving for Fhinge yields:
Substitute numerical values and
evaluate Fhinge:
Fhinge = Mg + mg − F = (M + m )g − F
(
)
Fhinge = (60 kg + 5.0 kg ) 9.81 m/s 2 − 181 N
= 0.46 kN
1198 Chapter 12
(c) The force diagram showing the
force F acting at right angles to the
board is shown to the right:
r
F
y
0m
1.5
r
Fhinge
0m
0.7
0m
0.8
θ
30°
P
Apply
∑τ
30°
r
Mg
r
mg
x
= 0 about the hinge:
F (3.0 m ) − mg [(1.5 m ) cos 30°] − Mg[(0.80 m ) cos 30°] = 0
Solving for F yields:
F=
m (1.5 m ) + M (0.80 m )
g cos 30°
3.0 m
Substitute numerical values and evaluate F:
F=
Apply
(5.0 kg )(1.5 m ) + (60 kg )(0.80 m ) (9.81 m/s 2 )cos 30° = 157 N =
3.0 m
∑F
y
0.16 kN
Fhinge sin θ − Mg − mg + F cos 30° = 0
= 0 to the board:
or
Fhinge sin θ = (M + m )g − F cos 30° (1)
Apply
∑F
x
Fhinge cos θ − F sin 30° = 0
= 0 to the board:
or
Fhinge cos θ = F sin 30°
Fhinge sin θ
Divide the first of these equations
by the second to obtain:
Fhinge cos θ
=
(M + m )g − F cos 30°
F sin 30°
⎡ (M + m )g − F cos 30° ⎤
⎥⎦
F sin 30°
⎣
Solving for θ yields:
θ = tan −1 ⎢
Substitute numerical values and evaluate θ :
(
)
⎡ (65 kg ) 9.81 m/s 2 − (157 N )cos30° ⎤
θ = tan ⎢
⎥ = 81.1°
(157 N )sin30°
⎣
⎦
−1
(2)
Static Equilibrium and Elasticity 1199
Substitute numerical values in
equation (2) and evaluate Fhinge:
Fhinge =
(157 N )sin 30° =
cos 81.1°
0.51kN
21 ••
A cylinder of mass M is supported by a frictionless trough formed by a
plane inclined at 30º to the horizontal on the left and one inclined at 60º on the right
as shown in Figure 12-35. Find the force exerted by each plane on the cylinder.
Picture the Problem The planes are
frictionless; therefore, the force exerted
by each plane must be perpendicular to
that plane. Let F1 be the force exerted by
the 30° plane, and let F2 be the force
exerted by the 60° plane. Choose a
coordinate system in which the positive
x direction is to the right and the positive
y direction is upward. Because the
cylinder is in equilibrium, we can use
the
conditions
for
translational
equilibrium to find the magnitudes of
F1 and F2 .
F1
r
Mg
30°
F2
60°
60°
30°
Apply
∑F
= 0 to the cylinder:
F1 sin 30° − F2 sin 60° = 0
Apply
∑F
= 0 to the cylinder:
F1 cos 30° + F2 cos 60° − Mg = 0
x
y
Solve equation (1) for F1:
Substitute for F1 in equation (2)
to obtain:
(1)
F1 = 3F2
(2)
(3)
3F2 cos 30° + F2 cos 60° − Mg = 0
Solve for F2 to obtain:
F2 =
Substitute for F2 in equation (3)
to obtain:
F1 = 3 ( 12 Mg ) =
Mg
=
3 cos 30° + cos 60°
3
2
1
2
Mg
Mg
22 ••
A uniform 18-kg door that is 2.0 m high by 0.80 m wide is hung from
two hinges that are 20 cm from the top and 20 cm from the bottom. If each hinge
supports half the weight of the door, find the magnitude and direction of the
horizontal components of the forces exerted by the two hinges on the door.
1200 Chapter 12
Picture the Problem The drawing
shows the door and its two supports. The
center of gravity of the door is 0.80 m
above (and below) the hinge, and 0.40 m
from the hinges horizontally. Choose a
coordinate system in which the positive
x direction is to the right and the positive
y direction is upward. Denote the
horizontal and vertical components of
the hinge force by FHh and FHv. Because
the door is in equilibrium, we can use
the conditions for translational and
rotational equilibrium to determine the
horizontal forces exerted by the hinges.
Apply
∑τ
= 0 about an axis
r
FHh
r
FHv
1.6 m
r
F'Hv
r
F'Hh
P
r
mg
0.40 m
FHh (1.6 m ) − mg (0.40 m ) = 0
through the lower hinge:
Solve for FHh:
Substitute numerical values and
evaluate FHh:
FHh =
FHh =
mg (0.40 m )
1.6 m
(18 kg ) (9.81 m/s2 )(0.40 m )
1.6 m
= 44 N
Apply
∑F
x
= 0 to the door and
solve for F 'Hh :
F ' Hh − FHh = 0
and
F 'Hh = 44 N
Remarks: Note that the upper hinge pulls on the door and the lower hinge
pushes on it.
23 ••
Find the force exerted on the strut by the hinge at A for the arrangement
in Figure 12-36 if (a) the strut is weightless, and (b) the strut weighs 20 N.
Static Equilibrium and Elasticity 1201
Picture the Problem Let T be the
tension in the line attached to the wall
and L be the length of the strut. The
figure includes w, the weight of the
strut, for part (b). Because the strut is
in equilibrium, we can use the
conditions for both rotational and
translational equilibrium to find the
force exerted on the strut by the hinge.
L
r
T
45°
1
0
r
Fv
45°
45°
A
(a) Express the force exerted on the
strut at the hinge:
Ignoring the weight of the strut,
apply ∑τ = 0 at the hinge:
Solve for the tension in the line:
Apply
∑ F = 0 to the strut:
r
W
L
2
r
w
r
Fh
F = Fh iˆ + Fv ˆj
(1)
LT − (L cos 45°)W = 0
T = W cos 45° = (60 N )cos 45°
= 42.4 N
∑F
x
= Fh − T cos 45° = 0
and
∑ Fy = Fv + T cos 45° − Mg = 0
Solve for and evaluate Fh:
Fh = T cos 45° = (42.4 N )cos 45° = 30 N
Solve for and evaluate Fv:
Fv = Mg − T cos 45°
Substitute in equation (1) to obtain:
F = (30 N ) iˆ + (30 N ) ˆj
(b) Including the weight of the strut,
apply ∑τ = 0 at the hinge:
⎞
⎛L
LT − (L cos 45°)W − ⎜ cos 45 ⎟ w = 0
⎠
⎝2
Solve for the tension in the line:
⎞
⎛1
T = (cos 45°)W + ⎜ cos 45° ⎟ w
⎠
⎝2
= 60 N − (42.4 N )cos45° = 30 N
1202 Chapter 12
⎛1
⎞
T = (cos 45°)(60 N ) + ⎜ cos 45° ⎟ (20 N )
⎝2
⎠
= 49.5 N
Substitute numerical values and
evaluate T:
Apply
∑F
∑ F = 0 to the strut:
x
= Fh − T cos 45° = 0
and
∑ Fy = Fv + T cos 45° − W − w = 0
Fh = T cos 45° = (49.5 N ) cos 45°
Solve for and evaluate Fh:
= 35 N
Fv = W + w − T cos 45°
Solve for and evaluate Fv:
= 60 N + 20 N − (49.5 N ) cos45°
= 45 N
F = (35 N ) iˆ + (45 N ) ˆj
Substitute for Fh and Fv to obtain:
24 ••
Julie has been hired to help paint the trim of a building, but she is not
convinced of the safety of the apparatus. A 5.0-m plank is suspended horizontally
from the top of the building by ropes attached at each end. Julie knows from
previous experience that the ropes being used will break if the tension exceeds
1.0 kN. Her 80-kg boss dismisses Julie’s worries and begins painting while standing
1.0 m from the end of the plank. If Julie’s mass is 60 kg and the plank has a mass of
20 kg, then over what range of positions can Julie stand to join her boss without
causing the ropes to break?
Picture the Problem Note that if Julie is at the far left end of the plank, T1 and T2
are less than 1.0 kN. Let x be the distance of Julie from T1. Because the plank is in
equilibrium, we can apply the condition for rotational equilibrium to relate the
distance x to the other distances and forces.
r
T2
r
T1
x
P
r
mJ g
1.5 m
r
mp g
1.0 m
r
mb g
Static Equilibrium and Elasticity 1203
Apply
∑τ
= 0 about an axis through the left end of the plank:
(5.0 m )T2 − (4.0 m ) mb g − (2.5 m ) mp g − mJ gx = 0
Solving for x yields:
x=
(5.0 m )T2 − (4.0 m ) mb − (2.5 m ) mp
mJ g
mJ
mJ
Substitute numerical values and
simplify to obtain:
m⎞
⎛
x = ⎜ 8.495 × 10−3 ⎟ T2 − 6.17 m
N⎠
⎝
Set T2 = 1.0 kN and evaluate x:
m⎞
⎛
x = ⎜ 8.495 × 10−3 ⎟ (1.0 kN ) − 6.17 m
N⎠
⎝
= 2.3 m
and Julie is safe provided x < 2.3 m.
25 ••
[SSM] A cylinder of mass M and radius R rolls against a step of height
h as shown in Figure 12-37. When a horizontal force of magnitude F is applied to
the top of the cylinder, the cylinder remains at rest. (a) Find an expression for the
normal force exerted by the floor on the cylinder. (b) Find an expression for the
horizontal force exerted by the edge of the step on the cylinder. (c) Find an
expression for the vertical component of the force exerted by the edge of the step on
the cylinder.
Picture the Problem The figure to
the right shows the forces acting on
the cylinder. Choose a coordinate
system in which the positive x
direction is to the right and the
positive y direction is upward.
Because the cylinder is in
equilibrium, we can apply the
conditions for translational and
rotational equilibrium to find Fn and
the
horizontal
and
vertical
components of the force the corner
of the step exerts on the cylinder.
(a) Apply
∑τ
= 0 to the cylinder
about the step’s corner:
r
F
R
R−h
r
Mg
r
Fc, v
r Fr
c, h
r
Fn
Mg − Fn − F (2 R − h ) = 0
h
1204 Chapter 12
Solving for Fn yields:
Express as a function of R and h:
Substitute for in the expression
for Fn and simplify to obtain:
Fn = Mg −
F (2 R − h )
= R 2 − (R − h ) = 2 Rh − h 2
2
Fn = Mg −
F (2 R − h )
2 Rh − h 2
= Mg − F
(b) Apply
∑F
x
= 0 to the cylinder:
2R − h
h
− Fc ,h + F = 0
Solve for Fc,h:
Fc ,h = F
(c) Apply ∑ Fy = 0 to the cylinder:
Fn − Mg + Fc ,v = 0 ⇒ Fc,v = Mg − Fn
Substitute the result from Part (a)
and simplify to obtain:
⎧
2R − h ⎫
Fc,v = Mg − ⎨Mg − F
⎬
h ⎭
⎩
= F
2R − h
h
26 ••
For the cylinder in Problem 25, find an expression for the minimum
magnitude of the horizontal force F that will roll the cylinder over the step if the
cylinder does not slide on the edge.
Picture the Problem The figure to the
right shows the forces acting on the
cylinder. Because the cylinder is in
equilibrium, we can use the condition
for rotational equilibrium to express
Fn in terms of F. Because, to roll over
the step, the cylinder must lift off the
floor, we can set Fn = 0 in our
expression relating Fn and F and solve
for F.
r
F
R
R−h
r
Mg
r
Fn
r
Fc, v
r Fr
c, h
h
Static Equilibrium and Elasticity 1205
Apply
∑τ
= 0 about the step’s
Mg − Fn − F (2 R − h ) = 0
corner:
Solve for Fn:
Express as a function of R and h:
Substitute for in the expression for
Fn and simplify to obtain:
Fn = Mg −
F (2 R − h )
= R 2 − (R − h ) = 2 Rh − h 2
2
Fn = Mg −
F (2 R − h )
2 Rh − h 2
= Mg − F
To roll over the step, the cylinder
must lift off the floor. That is,
Fn = 0:
Solving for F yields:
0 = Mg − F
F = Mg
2R − h
h
2R − h
h
h
2R − h
27 ••
Figure 12-38 shows a hand holding an epee, a weapon used in the sport
of fencing which you are taking as a physical education elective. The center of mass
of your epee is 24 cm from the pommel (the end of the epee at the grip). You have
weighed it so you know that the epee’s mass is 0.700 kg and its full length is
110 cm. (a) At the beginning of a match you hold it straight out in static
equilibrium. Find the total force exerted by your hand on the epee. (b) Find the
torque exerted by your hand on the epee. (c) Your hand, being an extended object,
actually exerts its force along the length of the epee grip. Model the total force
exerted by your hand as two oppositely directed forces whose lines of action are
separated by the width of your hand (taken to be 10.0 cm). Find the magnitudes and
directions of these two forces.
Picture the Problem The diagram shows the forces F1 and F2 that the fencer’s
hand exerts on the epee. We can use a condition for translational equilibrium to
find the upward force the fencer must exert on the epee when it is in equilibrium
and the definition of torque to determine the total torque exerted. In Part (c) we
can use the conditions for translational and rotational equilibrium to obtain two
equations in F1 and F2 that we can solve simultaneously. In Part (d) we can apply
Newton’s 2nd law in rotational form and the condition for translational
equilibrium to obtain two equations in F1 and F2 that, again, we can solve
simultaneously.
1206 Chapter 12
F2
12 cm
0
2 cm
F1
W
24 cm
(a) Letting the upward force exerted
by the fencer’s hand be F, apply
∑ Fy = 0 to the epee to obtain:
F −W = 0
and
F = W = mg
Substitute numerical values and
evaluate F:
F = (0.700 kg ) 9.81 m/s 2 = 6.87 N
(b) The torque due to the weight
about the left end of the epee is equal
in magnitude but opposite in
direction to the torque exerted by
your hand on the epee:
τ= w
Substitute numerical values and
evaluate τ :
τ = (0.24 m )(6.87 N ) = 1.65N ⋅ m
(c) Apply
∑F
y
= 0 to the epee to
(
)
= 1 .7 N ⋅ m
− F1 + F2 − 6.87 N = 0
(1)
obtain:
Apply
∑τ
0
= 0 to obtain:
− (0.020 m )F1 + (0.12 m )F2 − 1.65 N ⋅ m = 0
Solve equations (1) and (2)
simultaneously to obtain:
(2)
F1 = 8.3 N and F2 = 15 N .
Remarks: Note that the force nearest the butt of the epee is directed downward
and the force nearest the hand guard is directed upward.
28 ••
A large gate weighing 200 N is supported by hinges at the top and
bottom and is further supported by a wire as shown in Figure 12-39. (a) What must
be the tension in the wire for the force on the upper hinge to have no horizontal
component? (b) What is the horizontal force on the lower hinge? (c) What are the
vertical forces on the hinges?
Static Equilibrium and Elasticity 1207
Picture the Problem In the force diagram, the forces exerted by the hinges
are Fy , 2 , Fy ,1 , and Fx ,1 where the subscript 1 refers to the lower hinge. Because the
gate is in equilibrium, we can apply the conditions for translational and rotational
equilibrium to find the tension in the wire and the forces at the hinges.
r
T
θ
r
Fy ,2
r2
r
F y ,1
r
Fx ,1
r
1
(a) Apply
∑τ
= 0 about an axis
r
mg
T sin θ +
1
T cos θ − 1mg = 0
2
through the lower hinge and
perpendicular to the plane of the page:
Solving for T yields:
T=
Substitute numerical values and
evaluate T:
T=
mg
1 sin θ + 2 cos θ
1
(1.5 m )(200 N )
(1.5 m )sin 45° + (1.5 m )cos 45°
= 141 N = 0.14 kN
(b) Apply
∑F
x
= 0 to the gate:
Solve for and evaluate Fx,1:
Fx ,1 − T cos 45° = 0
Fx ,1 = T cos 45° = (141 N )cos 45°
= 99.7 N = 1.0 ×10 2 N
(c) Apply
∑F
y
= 0 to the gate:
Because Fy,1 and Fy,2 cannot be
determined independently, solve
for and evaluate their sum:
Fy ,1 + Fy , 2 + T sin 45° − mg = 0
Fy ,1 + Fy , 2 = mg − T sin 45°
= 200 N − 99.7 N
= 1.0 ×10 2 N
1208 Chapter 12
29 ••
On a camping trip, you moor your boat at the end of a dock in a rapidly
flowing river. It is anchored to the dock by a chain 5.0 m long, as shown in Figure
12-46. A 100-N weight is suspended from the center in the chain. This will allow
the tension in the chain to change as the force of the current which pulls the boat
away from the dock and to the right varies. The drag force by the water on the boat
depends on the speed of the water. You decide to apply the principles of statics you
learned in physics class. (Ignore the weight of the chain.) The drag force on the
boat is 50 N. (a) What is the tension in the chain? (b) How far is the boat from the
dock? (c) The maximum tension the chain can sustain is 500 N. What minimum
water drag force on the boat would snap the chain?
Picture the Problem The free-body diagram shown to the left below is for the
weight and the diagram to the right is for the boat. Because both are in
equilibrium under the influences of the forces acting on them, we can apply a
condition for translational equilibrium to find the tension in the chain.
y
y
Fn
T
T
θ
θ
x
θ
T
100 N
(a) Apply
Apply
∑F
x
∑F
y
= 0 to the boat:
= 0 to the weight:
Substitute for T to obtain:
Solve for θ to obtain:
Substitute for Fd and evaluate θ :
Solve equation (1) for T:
Substitute for θ and evaluate T:
x
Fd
mg
Fd − T cos θ = 0 ⇒ T =
Fd
cos θ
2T sin θ − 100 N = 0
2 Fd tan θ − 100 N = 0
⎡100 N ⎤
⎥
⎣ 2 Fd ⎦
θ = tan −1 ⎢
⎡ 100 N ⎤
⎥ = 45°
⎣ 2(50 N ) ⎦
θ = tan −1 ⎢
T=
100 N
2 sin θ
T=
100 N
= 70.7 N = 71 N
2 sin 45°
(1)
Static Equilibrium and Elasticity 1209
(b) Relate the distance d of the boat
from the dock to the angle θ the
chain makes with the horizontal:
cos θ =
Substitute numerical values and
evaluate d:
d = (5.0 m )cos 45° = 3.5 m
(c) Relate the resultant tension in the
chain to the vertical component of
the tension Fv and the maximum
drag force exerted on the boat by the
water Fd,max :
Fv2 + Fd,2max = (500 N )
Solve for Fd,max :
Fd,max =
(500 N )2 − Fv2
Because the vertical component of
the tension is 50 N:
Fd,max =
(500 N )2 − (50 N )2
1
2
1
2
d d
= ⇒ d = L cos θ
L L
2
= 0.50 kN
30 ••
Romeo takes a uniform 10-m ladder and leans it against the smooth
(frictionless) wall of the Capulet residence. The ladder’s mass is 22 kg and the
bottom rests on the ground 2.8 m from the wall. When Romeo, whose mass is
70 kg, gets 90 percent of the way to the top, the ladder begins to slip. What is the
coefficient of static friction between the ground and the ladder?
r
Fby wall
Picture the Problem The ladder and
the forces acting on it at the critical
moment of slipping are shown in the
diagram. Use the coordinate system
shown. Because the ladder is in
equilibrium, we can apply the
conditions for translational and
rotational equilibrium.
0.9r
y
0.5r
r
Fn
0
x
Using its definition, express µs:
r
μs =
10 m
r
m g Mgr
θ
r
f s,max
2.8 m
f s,max
Fn
(1)
1210 Chapter 12
Apply
∑τ
= 0 about the bottom of the ladder:
[0.9
cosθ ]Mg + [0.5 cosθ ]mg − [ sin θ ] FW = 0
Solving for FW yields:
Find the angle θ :
FW =
(0.9M + 0.5 m )g cosθ
sin θ
⎛ 2.8 m ⎞
⎟⎟ = 73.74°
10
m
⎝
⎠
θ = cos −1 ⎜⎜
Substitute numerical values and evaluate FW:
FW =
Apply
∑F
x
[0.9 (70 kg ) + 0.5 (22 kg )](9.81 m/s 2 )cos 73.74° = 211.7 N
= 0 to the ladder and
solve for fs,max:
Apply
∑F
y
= 0 to the ladder:
sin 73.74°
FW − f s, max = 0
and
f s,max = FW = 211.7 N
Fn − Mg − mg = 0 ⇒ Fn = (M + m )g
(
Substitute numerical values and
evaluate Fn:
Fn = (70 kg + 22 kg ) 9.81 m/s 2
Substitute numerical values in
equation (1) and evaluate µs:
μs =
)
= 902.5 N
211.7 N
= 0.23
902.5 N
31 ••
[SSM] Two 80-N forces are applied to opposite corners of a
rectangular plate as shown in Figure 12-41. (a) Find the torque produced by this
couple using Equation 12-6. (b) Show that the result is the same as if you determine
the torque about the lower left-hand corner.
Picture the Problem The forces shown in the figure constitute a couple and will
cause the plate to experience a counterclockwise angular acceleration. The couple
equation is τ = FD . The following diagram shows the geometric relationships
between the variables in terms of a generalized angle θ.
Static Equilibrium and Elasticity 1211
b− x
x
θ
θ
F = 80 N
D
a
a
F = 80 N
θ
r
P
b
(a) The couple equation is:
τ = FD
(1)
From the diagram, D is given by:
D = (b − x )cosθ
(2)
Again, referring to the diagram:
x = a tan θ
Substituting for x in equation (2)
and simplifying yields:
D = (b − a tan θ )cos θ
= b cos θ − a sin θ
Substituting for D in equation (1)
yields:
Substitute numerical values and
evaluate τ :
(b) Letting the counterclockwise
direction be the positive direction,
apply ∑τ = 0 about an axis normal
τ = F (b cosθ − a sin θ )
(3)
τ = (80 N )(b cos 30° − a sin 30°)
=
(69 N )b − (40 N ) a
F ( + D) − F = 0
to the plane of the rectangle and
passing through point P:
Substituting for D yields:
F ( + b cosθ − a sin θ ) − F = 0
Solve for τ to obtain:
τ = F (b cosθ − a sin θ ) , in agreement
with equation (3).
32 ••
A uniform cube of side a and mass M rests on a horizontal surface. A
horizontal force F is applied to the top of the cube as in Figure 12-42. This force is
not sufficient to move or tip the cube. (a) Show that the force of static friction
1212 Chapter 12
exerted by the surface and the applied force constitute a couple, and find the torque
exerted by the couple. (b) The torque exerted by the couple is balanced by the
torque exerted by the couple consisting of the normal force on the cube and the
gravitational force on the cube. Use this fact to find the effective point of
application of the normal force when F = Mg/3. (c) Find the greatest magnitude of
F for which the cube will not tip (Assuming the cube does not slip.).
Picture the Problem We can use the condition for translational equilibrium and
the definition of a couple to show that the force of static friction exerted by the
surface and the applied force constitute a couple. We can use the definition of
torque to find the torque exerted by the couple. We can use our result from (b) to
find the effective point of application of the normal force when F = Mg/3 and the
condition for rotational equilibrium to find the greatest magnitude of F for which
the cube will not tip.
(a) Apply
∑F
x
= 0 to the stationary
F + fs = 0 ⇒ F = − fs
cube:
Because F = − f s , this pair of equal, parallel, and oppositely directed forces
constitute a couple.
The torque of the couple is:
τ couple = Fa
(b) Let x equal the distance from the
point of application of Fn to the
center of the cube. Now, Fn = Mg, so
applying ∑τ = 0 to the cube yields:
Mgx − Fa = 0 ⇒ x =
Substituting for F and
simplifying yields:
Mg
a
a
3
x=
=
Mg
3
(c) Solve equation (1) for F:
F=
Noting that xmax = a/2, express the
condition that the cube will tip:
Fa
Mg
Mgx
a
Mgxmax
F>
=
a
a
2 = Mg
a
2
Mg
(1)
Static Equilibrium and Elasticity 1213
33 ••
[SSM] A ladder of negligible mass and of length L leans against a
slick wall making an angle of θ with the horizontal floor. The coefficient of friction
between the ladder and the floor is μs. A man climbs the ladder. What height h can
he reach before the ladder slips?
r
Fby wall
Picture the Problem Let the mass of
the man be M. The ladder and the
forces acting on it are shown in the
diagram. Because the wall is slick, the
force the wall exerts on the ladder must
be horizontal. Because the ladder is in
equilibrium, we can apply the
conditions for translational and
rotational equilibrium to it.
r
y
∑F
y
θ
r
Mg
r
Fn
0
Apply
L
θ
r
f s,max
= 0 to the ladder and
Fn − Mg = 0 ⇒ Fn = Mg
= 0 to the ladder and
f s,max − FW = 0 ⇒ f s,max = FW
h
x
solve for Fn:
Apply
∑F
x
solve for fs,max:
Apply
∑τ
= 0 about the bottom
Mg cosθ − FW L sin θ = 0
of the ladder to obtain:
Solving for and simplifying yields:
Referring to the figure, relate to
FW L sin θ f s,max L
=
tan θ
Mg cosθ
Mg
μFL
= s n tan θ = μ s L tan θ
Mg
=
h = sin θ
h:
Substituting for yields:
h = μ s L tan θ sin θ
34 ••
A uniform ladder of length L and mass m leans against a frictionless
vertical wall, making an angle of 60º with the horizontal. The coefficient of static
1214 Chapter 12
friction between the ladder and the ground is 0.45. If your mass is four times that of
the ladder, how high can you climb before the ladder begins to slip?
Picture the Problem The ladder and
L
the forces acting on it are shown in the
r
Fby wall
drawing. Choose a coordinate system in
which the positive x direction is to the
r
y
right and the positive y direction is
upward. Because the wall is smooth,
1
2 L
the force the wall exerts on the ladder
r
must be horizontal. Because the ladder
Fn
r
mg
is in equilibrium, we can apply the
r
4m g
conditions for translational and
r
θ
f s,max
rotational equilibrium.
0
x
Apply
∑F
y
= 0 to the ladder and
Fn − mg − 4mg = 0 ⇒ Fn = 5mg
= 0 to the ladder and
FW − f s,max = 0 ⇒ f s,max = FW
solve for Fn:
Apply
∑F
x
solve for f s,max :
Apply
∑τ
= 0 about an axis
mg
through the bottom of the ladder:
Substitute for FW and f s,max and
solve for :
Simplify to obtain:
Substitute numerical values to
obtain:
=
L
cos θ + 4mg cosθ − FW L sin θ = 0
2
5μs mgL sin θ − 12 mgL cos θ
4mg cos θ
1⎞
⎛ 5μ
= ⎜ s tan θ − ⎟ L
8⎠
⎝ 4
1⎞
⎛ 5(0.45)
=⎜
tan 60° − ⎟ L = 0.85 L
8⎠
⎝ 4
That is, you can climb about 85% of the
way to the top of the ladder before it
begins to slip.
A ladder of mass m and length L leans against a frictionless vertical
35 ••
wall, so that it makes an angle θ with the horizontal. The center of mass of the
ladder is a height h above the floor. A force F directed directly away from the wall
pulls on the ladder at its midpoint. Find the minimum coefficient of static friction μs
Static Equilibrium and Elasticity 1215
for which the top end of the ladder will separate from the wall before the lower end
begins to slip.
L
Picture the Problem The ladder and
the forces acting on it are shown in the
figure. Because the ladder is separating
from the wall, the force the wall exerts
on the ladder is zero. Because the
ladder is in equilibrium, we can apply
the conditions for translational and
rotational equilibrium.
y
r
F
1
2
L
r
Fn
0
To find the force required to pull the
ladder away from the wall, apply
∑τ = 0 about an axis through the
bottom of the ladder:
Solving for F yields:
r
mg
h
θ
r
f s,max
x
F ( 12 L sin θ ) − mg ( 12 L cosθ ) = 0
h
,
or, because 12 L cosθ =
tan θ
mgh
1
=0
2 LF sin θ −
tan θ
F=
2mgh
L tan θ sin θ
(1)
Apply
∑F
= 0 to the ladder:
f s,max − F = 0 ⇒ F = f s,max = μ s Fn (2)
Apply
∑F
= 0 to the ladder:
Fn − mg = 0 ⇒ Fn = mg
x
y
Equate equations (1) and (2) and
substitute for Fn to obtain:
Solving for µs yields:
μs mg =
μs =
2mgh
L tan θ sin θ
2h
L tan θ sin θ
36 ••
A 900-N man sits on top of a stepladder of negligible mass that rests on
a frictionless floor as in Figure 12-43. There is a cross brace halfway up the ladder.
The angle at the apex is θ = 30º. (a) What is the force exerted by the floor on each
leg of the ladder? (b) Find the tension in the cross brace. (c) If the cross brace is
moved down toward the bottom of the ladder (maintaining the same angle θ), will
its tension be the same, greater, or less than when it was in its higher position?
Explain your answer.
1216 Chapter 12
r
F
Picture the Problem Assume that half
the man’s weight acts on each side of the
ladder. The force exerted by the
frictionless floor must be vertical. D is
the separation between the legs at the
bottom and x is the distance of the cross
brace from the apex. Because each leg of
the ladder is in equilibrium, we can apply
the condition for rotational equilibrium to
the right leg to relate the tension in the
cross brace to its distance from the apex.
1
2
θ
1
2
r
wm
x
h
r
T
r
Fn
(a) By symmetry, each leg carries half the total weight, and the force on
each leg is 450 N.
D
FD
− Tx = 0 ⇒ T = n
2x
2
(b) Consider one of the ladder’s legs
and apply ∑τ = 0 about the apex:
Fn
Using trigonometry, relate h and θ
through the tangent function:
tan 12 θ =
Substitute for D in the expression for
T and simplify to obtain:
T=
Apply
∑F
y
= 0 to the ladder and
1
2
D
⇒ D = 2h tan 12 θ
h
2 Fn h tan 12 θ Fn h tan 12 θ
=
2x
x
Fn − 12 w = 0 ⇒ Fn = 12 w
solve for Fn:
Substitute for Fn to obtain:
T=
wh tan 12 θ
2x
Substitute numerical values and
evaluate T:
T=
(900 N )(4.0 m ) tan 15° =
2(2.0 m )
(1)
0.24 kN
(c) From equation (1) we can see that T is inversely proportional to x. Hence, if the
brace is moved lower, T will decrease.
37 ••
A uniform ladder rests against a frictionless vertical wall. The
coefficient of static friction between the ladder and the floor is 0.30. What is the
Static Equilibrium and Elasticity 1217
smallest angle between the ladder and the horizontal such that the ladder will not
slip?
r
Fby wall
Picture the Problem The figure shows
the forces acting on the ladder. Because
the wall is frictionless, the force the
wall exerts on the ladder is
perpendicular to the wall. Because the
ladder is on the verge of slipping, the
static friction force is fs,max. Because the
ladder is in equilibrium, we can apply
the conditions for translational and
rotational equilibrium.
r
y
1
2
r
mg
r
Fn
0
r
θ
r
f s,max
x
Apply
∑F
= 0 to the ladder:
f s,max − FW = 0 ⇒ FW = f s,max = μs Fn
Apply
∑F
= 0 to the ladder:
Fn − mg = 0 ⇒ Fn = mg
Apply
∑τ
x
y
= 0 about an axis
mg ( 12 cos θ ) − FW ( sin θ ) = 0
through the bottom of the ladder:
Substitute for FW and Fn and
simplify to obtain:
Substitute the numerical value of
μs and evaluate θ:
1
2
⎛ 1 ⎞
⎟⎟
cos θ − μs sin θ = 0 ⇒ θ = tan −1 ⎜⎜
⎝ 2 μs ⎠
⎡
1 ⎤
⎥ = 59°
⎣ 2(0.30) ⎦
θ = tan −1 ⎢
38 ••• A uniform log with a mass of 100 kg, a length of 4.0 m, and a radius of
12 cm is held in an inclined position, as shown in Figure 12-44. The coefficient of
static friction between the log and the horizontal surface is 0.60. The log is on the
verge of slipping to the right. Find the tension in the support wire and the angle the
wire makes with the vertical wall.
Picture the Problem Let T = the tension in the wire; Fn = the normal force of the
surface; and fs,max = µsFn the maximum force of static friction. Letting the point at
which the wire is attached to the log be the origin, the center of mass of the log is at
(−1.838 m, −0.797 m) and the point of contact with the floor is at
(−3.676 m, −1.594 m). Because the log is in equilibrium, we can apply the
conditions for translational and rotational equilibrium.
1218 Chapter 12
y
θ
r
T
r2
x
r
1
r3
r
Fn
r
f s,max
θ
r
mg
Apply
Apply
∑F
x
∑F
y
= 0 to the log:
= 0 to the log:
T sin θ − f s, max = 0
or
T sin θ = f s,max = μs Fn
(1)
T cosθ + Fn − mg = 0
or
T cos θ = mg − Fn
(2)
Divide equation (1) by equation (2)
to obtain:
μs Fn
T sin θ
=
T cos θ mg − Fn
Solving for θ yields:
⎞
⎛
⎟
⎜
μs ⎟
−1 ⎜
θ = tan
⎟
⎜ mg
− 1⎟
⎜
⎠
⎝ Fn
Apply
∑τ
= 0 about an axis through
2
mg − 1 Fn −
3
(3)
μs Fn = 0
the origin:
Solve for Fn to obtain:
Fn =
2
1
Substitute numerical values and
evaluate Fn:
Fn =
+
mg
3
μs
(
)
1.838(100 kg ) 9.81 m/s 2
= 389 N
3.676 + 1.594(0.60 )
Static Equilibrium and Elasticity 1219
⎤
⎡
⎥
⎢
0.60
⎥
θ = tan −1 ⎢
⎥
⎢ (100 kg ) 9.81m/s 2
− 1⎥
⎢
389 N
⎦
⎣
Substitute numerical values in
equation (3) and evaluate θ:
(
)
= 21.5° = 22°
Substitute numerical values in
equation (1) and evaluate T:
T=
(0.60)(389 N ) =
sin21.5°
0.64 kN
39 ••• [SSM] A tall, uniform, rectangular block sits on an inclined plane as
shown in Figure 12-45. A cord is attached to the top of the block to prevent it from
falling down the incline. What is the maximum angle θ for which the block will not
slide on the incline? Assume the block has a height-to-width ratio, b/a, of 4.0 and
the coefficient of static friction between it and the incline is μs = 0.80.
Picture the Problem Consider what happens just as θ increases beyond θ max.
Because the top of the block is fixed by the cord, the block will in fact rotate with
only the lower right edge of the block remaining in contact with the plane. It
follows that just prior to this slipping, Fn and fs = µsFn act at the lower right edge of
the block. Choose a coordinate system in which up the incline is the +x direction
and the direction of Fn is the +y direction. Because the block is in equilibrium, we
can apply the conditions for translational and rotational equilibrium.
y
a
b
r
T
θ
+
r
Fn
r
mg
r
f s,max
θ
x
θ
Apply
∑F
= 0 to the block:
T + μs Fn − mg sin θ = 0
(1)
Apply
∑F
= 0 to the block:
Fn − mg cos θ = 0
(2)
Apply
∑τ
x
y
= 0 about an axis through
the lower right edge of the block:
1
2
a(mg cosθ ) + 12 b(mg sin θ ) − bT = 0 (3)
1220 Chapter 12
Eliminate Fn between equations
(1) and (2) and solve for T:
T = mg (sin θ − μs cos θ )
Substitute for T in equation (3):
1
2
a(mg cosθ ) + 12 b(mg sin θ )
Substitute 4a for b:
1
2
a(mg cosθ ) + 12 (4.0a )(mg sin θ )
Simplify to obtain:
(1 + 8.0μs ) cos θ − 4.0 sin θ = 0
Solving for θ yields:
Substitute numerical values and
evaluate θ :
− b[mg (sin θ − μ s cosθ )] = 0
− (4.0a )[mg (sin θ − μ s cosθ )] = 0
⎡1 + 8.0μs ⎤
⎥
⎣ 4.0 ⎦
θ = tan −1 ⎢
⎡1 + (8.0)(0.80) ⎤
⎥ = 62°
4.0
⎣
⎦
θ = tan −1 ⎢
Stress and Strain
40 •
A 50-kg ball is suspended from a steel wire of length 5.0 m and radius
2.0 mm. By how much does the wire stretch?
Picture the Problem L is the unstretched length of the wire, F is the force acting
on it, and A is its cross-sectional area. The stretch in the wire ΔL is related to
Young’s modulus by Y = (F A) (ΔL L ). We can use Table 12-1 to find the
numerical value of Young’s modulus for steel.
FL
F A
⇒ ΔL =
ΔL L
YA
Find the amount the wire is
stretched from Young’s modulus:
Y=
Substitute for F and A to obtain:
ΔL =
Substitute numerical values and
evaluate ΔL:
(
50 kg ) (9.81 m/s 2 )(5.0 m )
ΔL =
(200 GN/m 2 )(π )(2.0 ×10−3 m )2
mgL
Yπ r 2
= 0.98 mm
41 •
[SSM] Copper has a tensile strength of about 3.0 × 108 N/m2. (a) What
is the maximum load that can be hung from a copper wire of diameter 0.42 mm?
Static Equilibrium and Elasticity 1221
(b) If half this maximum load is hung from the copper wire, by what percentage of
its length will it stretch?
Picture the Problem L is the unstretched length of the wire, F is the force acting
on it, and A is its cross-sectional area. The stretch in the wire ΔL is related to
Young’s modulus by Y = stress strain = (F A) (ΔL L ).
(a) Express the maximum load in
terms of the wire’s tensile strength:
Fmax = tensile strength × A
Substitute numerical values and
evaluate Fmax:
Fmax = 3.0 × 108 N/m 2 π 0.21× 10 −3 m
(b) Using the definition of Young’s
modulus, express the fractional
change in length of the copper wire:
1
F
ΔL
F
=
= 2 max
L
AY
AY
Substitute numerical values and
ΔL
:
evaluate
L
= tensile strength × π r 2
(
) (
)
2
= 41.6 N = 42 N
1
ΔL
2 (41.6 N )
=
L π (0.21 mm )2 (1.10 × 1011 N/m 2 )
= 0.14%
42 •
A 4.0-kg mass is supported by a steel wire of diameter 0.60 mm and
length 1.2 m. How much will the wire stretch under this load?
Picture the Problem L is the unstretched length of the wire, F is the force acting on
it, and A is its cross-sectional area. The stretch in the wire ΔL is related to Young’s
modulus by Y = (F A) (ΔL L ). We can use Table 12-1 to find the numerical value
of Young’s modulus for steel.
FL
F A
⇒ ΔL =
ΔL L
YA
Relate the amount the wire is
stretched to Young’s modulus:
Y=
Substitute for F and A to obtain:
ΔL =
Substitute numerical values and
evaluate ΔL:
ΔL =
mgL
Yπ r 2
(4.0 kg ) (9.81 m/s 2 )(1.2 m )
(
2π × 1011 N/m 2 0.30 ×10 −3 m
= 0.83 mm
)
2
1222 Chapter 12
43 •
[SSM] As a runner’s foot pushes off on the ground, the shearing force
acting on an 8.0-mm-thick sole is shown in Figure 12-46. If the force of 25 N is
distributed over an area of 15 cm2, find the angle of shear θ, given that the shear
modulus of the sole is 1.9 × 105 N/m2.
Picture the Problem The shear stress, defined as the ratio of the shearing force to
the area over which it is applied, is related to the shear strain through the definition
shear stress Fs A
=
.
of the shear modulus; M s =
shear strain tan θ
Using the definition of shear modulus,
relate the angle of shear, θ to the shear
force and shear modulus:
Substitute numerical values and
evaluate θ :
tan θ =
⎛ F ⎞
Fs
⇒ θ = tan −1 ⎜⎜ s ⎟⎟
Ms A
⎝ MsA⎠
⎤
25 N
2
−4
2 ⎥
⎣ 1.9 ×10 N/m 15 × 10 m ⎦
⎡
θ = tan −1 ⎢
(
5
)(
)
= 5 .0 °
A steel wire of length 1.50 m and diameter 1.00 mm is joined to an
44 ••
aluminum wire of identical dimensions to make a composite wire of length 3.00 m.
Find the resulting change in the length of this composite wire if an object with a
mass of 5.00 kg is hung vertically from one of its ends. (Neglect any effects the
masses of the two wires have on the changes in their lengths.)
Picture the Problem The stretch in the wire ΔL is related to Young’s modulus
by Y = (F A) (ΔL L ) , where L is the unstretched length of the wire, F is the force
acting on it, and A is the cross-sectional area of the wire. The change in length of
the composite wire is the sum of the changes in length of the steel and aluminum
wires.
The change in length of the
composite wire ΔL is the sum of the
changes in length of the two wires:
ΔL = ΔLsteel + ΔLAl
Static Equilibrium and Elasticity 1223
Using the defining equation for
Young’s modulus, substitute for
ΔLsteel and ΔLAl in equation (1) and
simplify to obtain:
ΔL =
F Lsteel F LAl
+
A Ysteel A YAl
=
F ⎛ Lsteel LAl ⎞
⎜
⎟
+
A ⎜⎝ Ysteel YAl ⎟⎠
Substitute numerical values and evaluate ΔL:
ΔL =
(5.00 kg ) (9.81 m/s 2 ) ⎛
(
π 0.500 × 10 −3 m
)
2
1.50 m
1.50 m
⎞
= 1.81× 10 −3 m
+
⎜
11
2
11
2 ⎟
0.700 ×10 N/m ⎠
⎝ 2.00 ×10 N/m
= 1.81 mm
45 ••
[SSM] Equal but opposite forces of magnitude F are applied to both
ends of a thin wire of length L and cross-sectional area A. Show that if the wire is
modeled as a spring, the force constant k is given by k = AY/L and the potential
energy stored in the wire is U = 12 FΔL , where Y is Young’s modulus and ΔL is the
amount the wire has stretched.
Picture the Problem We can use Hooke’s law and Young’s modulus to show that,
if the wire is considered to be a spring, the force constant k is given by k = AY/L. By
treating the wire as a spring we can show the energy stored in the wire is
U = 12 FΔL .
Express the relationship between the
stretching force, the force constant,
and the elongation of a spring:
F = kΔL ⇒ k =
Using the definition of Young’s
modulus, express the ratio of the
stretching force to the elongation of
the wire:
F
AY
=
ΔL
L
F
ΔL
(1)
Equate these two expressions for
F/ΔL to obtain:
k=
Treating the wire as a spring, express
its stored energy:
U = 12 k (ΔL ) =
AY
L
2
1
2
AY
(ΔL ) 2
L
⎛ AYΔL ⎞
= 12 ⎜
⎟ΔL
⎝ L ⎠
1224 Chapter 12
Solving equation (1) for F yields:
F=
Substitute for F in the expression for
U to obtain:
U=
AYΔL
L
1
2
FΔL
46 ••
The steel E string of a violin is under a tension of 53.0 N. The diameter
of the string is 0.200 mm and the length under tension is 35.0 cm. Find (a) the
unstretched length of this string and (b) the work needed to stretch the string.
Picture the Problem Let L′ represent the stretched and L the unstretched length of
the wire. The stretch in the wire ΔL is related to Young’s modulus
by Y = (F A) (ΔL L ) , where F is the force acting on it, and A is its cross-sectional
area. In Problem 45 we showed that the energy stored in the wire is U = 12 FΔL ,
where Y is Young’s modulus and ΔL is the amount the wire has stretched.
(a) Express the stretched length L′
of the wire:
L' = L + ΔL
Using the definition of Young’s
modulus, express ΔL:
ΔL =
Substitute and simplify:
L' = L +
Solving for L yields:
L=
LF
AY
LF
F ⎞
⎛
= L⎜1 +
⎟
AY
⎝ AY ⎠
L'
F
1+
AY
Substitute numerical values and evaluate L:
L=
1+
(
0.350 m
53.0 N
π 0.100 × 10 −3 m
= 34.7 cm
) (2.00 ×10
2
11
N/m 2
)
(b) From Problem 45, the work done
in stretching the wire is:
W = ΔU = 12 FΔL
Substitute numerical values and
evaluate W:
W=
1
2
(53.0 N )(0.350 m − 0.347 m )
≈ 0.08 J
Static Equilibrium and Elasticity 1225
47 ••
During a materials science experiment on the Young’s modulus of
rubber, your teaching assistant supplies you and your team with a rubber strip that
is rectangular in cross section. She tells you to first measure the cross section
dimensions and their values are 3.0 mm × 1.5 mm. The lab write-up calls for the
rubber strip to be suspended vertically and various (known) masses to attached to it.
Your team obtains the following data for the length of the strip as a function of the
load (mass) on the end of the strip:
Load, kg
0.0 0.10 0.20 0.30 0.40 0.50
Length, cm 5.0 5.6 6.2 6.9 7.8 8.8
(a) Use a spreadsheet or graphing calculator to find Young’s modulus for the
rubber strip over this range of loads. Hint: It is probably best to plot F/A
versus ΔL/L. Why?
(b) Find the energy stored in the strip when the load is 0.15 kg. (See Problem 45.)
(c) Find the energy stored in the strip when the load is 0.30 kg. Is it twice as
much as your answer to Part (b)? Explain.
Picture the Problem We can use the definition of Young’s modulus and your
team’s data to plot a graph whose slope is Young’s modulus for the rubber strip
over the given range of loads. Because the rubber strip stretches linearly for loads
less than or equal to 0.20 kg, we can use linear interpolation in Part (b) to find the
length of the rubber strip for a load of 0.15 kg. We can then use the result of
Problem 45 to find the energy stored in the when the load is 0.15 kg. In Part (c) we
can use the result of Problem 45 and the given length of the strip when its load is
0.30 kg to find the energy stored in the rubber strip.
F
ΔL
=Y
A
L
where Y is the slope of a graph of F/A
as a function of ΔL/L.
(a) The equation for Young’s
modulus can be written as:
The following table summarizes the quantities, calculated using your team’s data,
used to plot the graph suggested in the problem statement.
Load
(kg)
0.10
0.20
0.30
0.40
0.50
F
(N)
0.981
1.962
2.943
3.924
4.905
F/A
(N/m2)
21.8×105
4.36×105
6.54×105
8.72×105
10.9×105
ΔL ΔL/L
U
(m)
(J)
0.006 0.12 2.9×10−3
0.012 0.24 12×10−3
0.019 0.38 28×10−3
0.028 0.56 55×10−3
0.038 0.76 93×10−3
1226 Chapter 12
A spreadsheet-generated graph of F/A as a function of ΔL/L follows. The
spreadsheet program also plotted the regression line on the graph and added its
equation to the graph.
1.2E+06
F /A , N/m^2
1.0E+06
8.0E+05
6.0E+05
y = 1.35E+06x + 9.89E+04
4.0E+05
2.0E+05
0.0E+00
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
delta-L /L
From the regression function shown
on the graph:
Y = 1.4 × 106 N/m 2
(b) From Problem 45:
U = 12 ΔLF
or, because F = mg,
U (m ) = 12 ΔLmg
Interpolating from the data table we see that the length of the strip when the
load on it is 0.15 kg is 5.9 cm. Substitute numerical values and evaluate
U(0.15 kg):
U (0.30 kg ) =
1
2
(5.9 cm − 5.0 cm)(0.15 kg )(9.81 m/s 2 ) =
7 mJ
(c) Evaluate U(0.30 kg) to obtain:
U (0.30 kg ) =
1
2
(6.9 cm − 5.0 cm)(0.30 kg )(9.81 m/s 2 ) =
28 mJ
The energy stored in the strip when the load is 0.30 kg is four times as much as the
energy stored when the load is 0.15 kg. Although the rubber strip does not stretch
linearly (a conclusion you can confirm either graphically or by examining the data
table), its stretch is sufficiently linear that, to a good approximation, the energy
stored is quadrupled when the load is doubled.
Static Equilibrium and Elasticity 1227
48 ••
A large mirror is hung from a nail as shown in Figure 12-47. The
supporting steel wire has a diameter of 0.20 mm and an unstretched length of 1.7 m.
The distance between the points of support at the top of the mirror’s frame is 1.5 m.
The mass of the mirror is 2.4 kg. How much will the distance between the nail and
the mirror increase due to the stretching of the wire as the mirror is hung?
Picture the Problem The figure shows
the forces acting on the wire where it
passes over the nail. m represents the
mass of the mirror and T is the tension
in the supporting wires. The figure also
shows the geometry of the right triangle
defined by the support wires and the top
of the mirror frame. The distance a is
fixed by the geometry while h and L will
change as the mirror is suspended from
the nail.
y
r
r
Fby nail = m g
nail
r
T'
θ θ
h
L=
r
T
x
0.8
5m
a = 0.75 m
Using the Pythagorean theorem,
express the relationship between the
sides of the right triangle in the
diagram:
a 2 + h 2 = L2
Express the differential of this
equation and approximate
differential changes with small
changes:
2aΔa + 2hΔh = 2 LΔL
or, because Δa = 0,
LΔL
hΔh = LΔL ⇒ Δh =
h
Multiplying the numerator and
denominator by L yields:
Δh =
Solve the equation defining Young’s
modulus for ΔL/L to obtain:
ΔL T
=
L
AY
Substitute for ΔL/L in equation (1) to
obtain:
T
(2)
2
L2 − a 2 π r Y
where r is the radius of the wire.
Noting that T = T' , apply
∑F
y
=0
to the wire where it passes over the
supporting nail:
Δh =
L2 ΔL
h L
L2 T
=
h AY
(1)
L2
mg − 2T cosθ = 0 ⇒ T =
mg
2 cosθ
1228 Chapter 12
Substituting for T in equation (2)
yields:
h
Because cosθ = =
L
L2 − a 2
:
L
Δh =
Δh =
=
L2
mg
L − a 2π r Y cosθ
2
2
2
L2
mgL
L − a 2π r Y L2 − a 2
2
2
2
mgL3
2π r 2Y L2 − a 2
(
)
Substitute numerical values and evaluate Δh:
Δh =
(2.4 kg )(9.81 m/s 2 )(0.85 m )3
=
2
2
2
2π (0.10 × 10−3 m ) (2.00 × 1011 N/m 2 )[(0.85 m ) − (0.75 m ) ]
7.2 mm
49 ••
Two masses, M1 and M2, are supported by wires that have equal lengths
when unstretched. The wire supporting M1 is an aluminum wire 0.70 mm in
diameter, and the one supporting M2 is a steel wire 0.50 mm in diameter. What is
the ratio M1/M2 if the two wires stretch by the same amount?
Picture the Problem Let the numeral 1 denote the aluminum wire and the
numeral 2 the steel wire. Because their initial lengths and amount they stretch are
the same, we can use the definition of Young’s modulus to express the change in
the lengths of each wire and then equate these expressions to obtain an equation
solvable for the ratio M1/M2.
Using the definition of Young’s
modulus, express the change in
length of the aluminum wire:
ΔL1 =
M 1 gL1
A1YAl
Using the definition of Young’s
modulus, express the change in
length of the steel wire:
ΔL2 =
M 2 gL2
A2Ysteel
Because the two wires stretch by the
same amount, equate ΔL1 and ΔL2
and simplify:
M1
M2
M
AY
=
⇒ 1 = 1 Al
M 2 A2Ysteel
A1YAl A2Ysteel
Static Equilibrium and Elasticity 1229
Substitute numerical values and
evaluate M1/M2:
π
(0.70 mm)2 (0.70 × 1011 N/m 2 )
M1 4
=
2
M2 π (
0.50 mm ) 2.00 × 1011 N/m 2
4
(
)
= 0.69
50 ••
A 0.50-kg ball is attached to one end of an aluminum wire having a
diameter of 1.6 mm and an unstretched length of 0.70 m. The other end of the wire
is fixed to the top of a post. The ball rotates about the post in a horizontal plane at a
rotational speed such that the angle between the wire and the horizontal is 5.0º. Find
the tension in the wire and the increase in its length due to the tension in the wire.
Picture the Problem The free-body
diagram shows the forces acting on the
ball as it rotates around the post in a
horizontal plane. We can apply
Newton’s 2nd law to find the tension in
the wire and use the definition of
Young’s modulus to find the amount by
which the aluminum wire stretches.
Apply
∑F
y
= 0 to the ball:
Substitute numerical values and
evaluate T:
x
y
r
T
θ
m
r
mg
T sin θ − mg = 0 ⇒ T =
mg
sin θ
(
0.50 kg ) (9.81 m/s 2 )
T=
= 56.3 N
sin5.0°
= 56 N
Using the definition of Young’s
modulus, express ΔL:
ΔL =
Substitute numerical values and
evaluate ΔL:
ΔL =
FL
AY
π
(56.3 N )(0.70 m )
(1.6 ×10
4
−3
m
) (0.70 ×10
2
11
N/m 2
)
= 0.28 mm
[SSM] An elevator cable is to be made of a new type of composite
51 ••
developed by Acme Laboratories. In the lab, a sample of the cable that is 2.00 m
long and has a cross-sectional area of 0.200 mm2 fails under a load of 1000 N. The
actual cable used to support the elevator will be 20.0 m long and have a cross-
1230 Chapter 12
sectional area of 1.20 mm2. It will need to support a load of 20,000 N safely. Will
it?
Picture the Problem We can use the definition of stress to calculate the failing
stress of the cable and the stress on the elevator cable. Note that the failing stress
of the composite cable is the same as the failing stress of the test sample.
The stress on the elevator cable is:
Stress cable =
F
20.0 kN
=
A 1.20 × 10 −6 m 2
= 1.67 ×1010 N/m 2
The failing stress of the sample is:
Stress failing =
F
1000 N
=
A 0.2 × 10 −6 m 2
= 0.500 × 1010 N/m 2
Because Stress failing < Stress cable , the cable will not support the elevator.
52 ••
If a material’s density remains constant when it is stretched in one
direction, then (because its total volume remains constant), its length must decrease
in one or both of the other directions. Take a rectangular block of length x, width y,
and depth z, and pull on it so that its new length x' = x + Δx . If Δx << x and
Δy y = Δz z , show that Δy y = − 12 Δx x .
Picture the Problem Let the length of the sides of the rectangle be x, y and z.
Then the volume of the rectangle will be V = xyz and we can express the new
volume V ′ resulting from the pulling in the x direction and the change in volume
ΔV in terms of Δx, Δy, and Δz. Discarding the higher order terms in ΔV and
dividing our equation by V and using the given condition that Δy/y = Δz/z will lead
us to the given expression for Δy/y.
Express the new volume of the rectangular box when its sides change in length by
Δx, Δy, and Δz:
V' = (x + Δx )( y + Δy )(z + Δz ) = xyz + Δx( yz ) + Δy (xz ) + Δz (xy )
+ {zΔxΔy + yΔxΔz + xΔyΔz + ΔxΔyΔz}
where the terms in brackets are very small (i.e., second order or higher).
Discard the second order and higher
terms to obtain:
Because ΔV = 0:
V' = V + Δx( yz ) + Δy ( xz ) + Δz ( xy )
or
ΔV = V' − V = Δx( yz ) + Δy ( xz ) + Δz ( xy )
Δx( yz ) = −[Δy ( xz ) + Δz ( xy )]
Static Equilibrium and Elasticity 1231
Divide both sides of this equation by
V = xyz to obtain:
⎡ Δy Δz ⎤
Δx
= −⎢ + ⎥
x
z ⎦
⎣ y
Because Δy/y = Δz/z, our equation
becomes:
Δy
1 Δx
Δx
Δy
⇒
= −
= −2
x
y
y
2 x
53 ••
[SSM] You are given a wire with a circular cross-section of radius r
and a length L. If the wire is made from a material whose density remains constant
when it is stretched in one direction, then show that Δr r = − 12 ΔL L , assuming that
ΔL << L . (See Problem 52.)
Picture the Problem We can evaluate the differential of the volume of the wire
and, using the assumptions that the volume of the wire does not change under
stretching and that the change in its length is small compared to its length, show
that Δr r = − 12 ΔL L L.
Express the volume of the wire:
V = π r2L
Evaluate the differential of V to
obtain:
dV = π r 2 dL + 2π rLdr
Because dV = 0:
Because ΔL << L, we can
approximate the differential changes
dr and dL with small changes Δr and
ΔL to obtain:
0 = rdL + 2 Ldr ⇒
1 dL
dr
=−
r
2 L
Δr
1 ΔL
= −
r
2 L
54 ••• For most materials listed in Table 12-1, the tensile strength is two to
three orders of magnitude lower than Young’s modulus. Consequently, most of
these materials will break before their strain exceeds 1 percent. Of man-made
materials, nylon has about the greatest extensibility—it can take strains of about 0.2
before breaking. But spider silk beats anything man-made. Certain forms of spider
silk can take strains on the order of 10 before breaking! (a) If such a thread has a
circular cross-section of radius r0 and unstretched length L0, find its new radius r
when stretched to a length L = 10L0. (Assume that the density of the thread remains
constant as it stretches.) (b) If the Young’s modulus of the spider thread is Y,
calculate the tension needed to break the thread in terms of Y and r0.
Picture the Problem Because the density of the thread remains constant during
the stretching process, we can equate the initial and final volumes to express r0 in
terms of r. We can also use Young’s modulus to express the tension needed to
break the thread in terms of Y and r0.
1232 Chapter 12
(a) Because the volume of the thread
is constant during the stretching of
the spider’s silk:
π r 2 L = π r02 L0 ⇒ r = r0
Substitute for L and simplify to
obtain:
r = r0
(b) Express Young’s modulus in
terms of the breaking tension T:
Y=
T A T π r 2 10T π r02
=
=
ΔL L ΔL L
ΔL L
T=
1
ΔL
π r02Y
10
L
Solving for T yields:
Because ΔL/L = 9:
L0
L
L0
= 0.316r0
10 L0
9π r02Y
T=
10
General Problems
55 •
[SSM] A standard bowling ball weighs 16 pounds. You wish to hold a
bowling ball in front of you, with the elbow bent at a right angle. Assume that your
biceps attaches to your forearm at 2.5 cm out from the elbow joint, and that your
biceps muscle pulls vertically upward, that is, it acts at right angles to the forearm.
Also assume that the ball is held 38 cm out from the elbow joint. Let the mass of
your forearm be 5.0 kg and assume its center of gravity is located 19 cm out from
the elbow joint. How much force must your biceps muscle apply to forearm in order
to hold out the bowling ball at the desired angle?
Picture the Problem We can model the forearm as a cylinder of length L = 38 cm
with the forces shown in the pictorial representation acting on it. Because the
forearm is in both translational and rotational equilibrium under the influence of
these forces, the forces in the diagram must add (vectorially) to zero and the net
torque with respect to any axis must also be zero.
r
Fbicep
1
r
L
0
2 L
x
elbow
r
r
mforearm g
Felbow joint
r
m ball g
Static Equilibrium and Elasticity 1233
Apply
∑τ = 0 about an axis through
Fbicep − 12 Lmforearm g − Lm ball g = 0
the elbow and perpendicular to the
plane of the diagram:
Solving for Fbicep and simplifying
Fbicep =
yields:
=
1
2
Lmforearm g + Lmball g
( 12 mforearm + mball )Lg
Substitute numerical values and evaluate Fbicep :
(
Fbicep
)
1 kg ⎞
⎛1
2
⎟(38 cm ) 9.81 m/s
⎜ 2 (5.0 kg ) + 16 lb ×
2.205 lb ⎠
= 1.5 kN
=⎝
2.5 cm
56 ••
A biology laboratory at your university is studying the location of a
person’s center of gravity as a function of their body weight. They pay well, and
you decide to volunteer. The location of your center of gravity when standing erect
is to be determined by having you lie on a uniform board (mass of 5.00 kg, length
2.00 m) supported by two scales as shown in Figure 12-54. If your height is 188 cm
and the left scale reads 470 N while the right scale reads 430 N, where is your
center of gravity relative to your feet? Assume the scales are both exactly the same
distance from the two ends of the board, are separated by 178 cm, and are set to
each read zero before you get on the platform.
Picture the Problem Because the you-board system is in equilibrium, we can apply
the conditions for translational and rotational equilibrium to relate the forces
exerted by the scales to the distance d, measured from your feet, to your center of
mass and the distance to the center of gravity of the board. The following pictorial
representation shows the forces acting on the board. mg represents your weight.
470 N
430 N
178 cm
d
P
89 cm
(5.00 kg )g
mg
1234 Chapter 12
The forces responsible for a counterclockwise torque about an axis through your
feet (point P) and perpendicular to the page are your weight and the weight of the
board. The only force causing a clockwise torque about this axis is the 470 N force
exerted by the scale under your head. Apply ∑τ = 0 about an axis through your
feet and perpendicular to the page:
(
)
m 9.81 m/s 2 d − (5.0 kg )(0.89 m ) − (1.78 m )(470 N ) = 0
where m is your mass.
Solve for d to obtain:
d=
(5.00 kg )(0.89 m ) + (1.78 m )(470 N )
m (9.81 m/s 2 )
Let upward be the positive y direction and apply
(
)
∑F
y
(1)
= 0 to the plank to obtain:
(
)
470 N + 430 N − m 9.81 m/s 2 − (5.00 kg ) 9.81 m/s 2 = 0
m = 86.74 kg
Solving for m yields:
Substitute numerical values in equation (1) and evaluate d:
d=
(5.0 kg )(0.89 m ) + (1.78 m )(470 N ) =
(86.74 kg )(9.81 m/s2 )
99 cm
Figure 12-49 shows a mobile consisting of four objects hanging on three
57 ••
rods of negligible mass. Find the values of the unknown masses of the objects if the
mobile is to balance. Hint: Find the mass m1 first.
Picture the Problem We can apply the balance condition
∑τ
= 0 successively,
starting with the lowest part of the mobile, to find the value of each of the unknown
weights.
Apply
∑τ
= 0 about an axis through
(3.0 cm )(2.0 N ) − (4.0 cm )m1 g = 0
the point of suspension of the lowest
part of the mobile:
Solving for m1 yields:
m1 = 0.1529 kg = 0.15 kg
Static Equilibrium and Elasticity 1235
Apply
∑τ
= 0 about an axis through the point of suspension of the middle part
of the mobile:
(2.0 cm ) m2 g − (4.0 cm )⎛⎜⎜ 2.0 N + 0.1529 kg ⎞⎟⎟ g = 0
g
⎝
Solving for m2 yields:
Apply
∑τ
⎠
m2 = 0.7136 kg = 0.71 kg
= 0 about an axis through the point of suspension of the top part of
the mobile:
(2.0 cm )(2.0 N + (0.7136 kg )g + (0.1529)g ) − (6.0 cm )m3 g = 0
Solving for m3 yields:
m3 = 0.3568 kg = 0.36 kg
58 ••
Steel construction beams, with an industry designation of ″W12 × 22,″
have a weight of 22 pounds per foot. A new business in town has hired you to place
its sign on a 4.0 m long steel beam of this type. The design calls for the beam to
extend outward horizontally from the front brick wall (Figure 12-50). It is to be held
in place by a 5.0 m-long steel cable. The cable is attached to one end of the beam
and to the wall above the point at which the beam is in contact with the wall.
During the initial stage of construction, the beam is not to be bolted to the wall, but
to be held in place solely by friction. (a) What is the minimum coefficient of
friction between the beam and the wall for the beam to remain in static equilibrium?
(b)What is the tension in the cable in this case?
Picture the Problem Because the beam is in both translational and rotational
equilibrium under the influence of the forces shown below in the pictorial
representation, we can apply ∑ F = 0 and ∑τ = 0 to it to find the coefficient of
static friction and the tension in the supporting cable.
r
T
r
fs
0
1
2
L
r
Fn
r
m beam g
θ
L
x
1236 Chapter 12
(a) Apply
∑ F = 0 to the beam to
∑F
x
= Fn − T cos θ = 0
and
∑ Fy = f s − mbeam g + T sin θ = 0
obtain:
The mass of the beam mbeam is the
(1)
(2)
mbeam = λL
product of its linear density λ and
length L:
Substituting for mbeam in equation (2)
f s − λLg + T sin θ = 0
(3)
yields:
Relate the force of static friction to
the normal force exerted by the wall:
f s = μs Fn
Substituting for fs in equation (3)
yields:
μs Fn − λLg + T sin θ = 0
Solve for μs to obtain:
μs =
λLg − T sin θ
Fn
Solving equation (1) for Fn yields:
Fn = T cos θ
Substitute for Fn in equation (4) and
simplify to obtain:
μs =
Apply
∑τ = 0 to the beam about an
(4)
λ Lg − T sin θ
λ Lg
=
− tan θ (5)
T cos θ
T cos θ
(T sin θ )L − mbeam g ( 12 L) = 0
axis through the origin and normal to
the page to obtain:
or
(T sin θ )L − λLg ( 12 L) = 0
Solving for T yields:
T=
Substitute for T in equation (5) and
simplify to obtain:
μs =
From Figure 12-50 we see that:
tan θ =
λLg
2 sin θ
λLg
λLg
cosθ
2 sin θ
3.0 m 3
=
4.0 m 4
(6)
− tan θ = tan θ
(7)
Static Equilibrium and Elasticity 1237
Substituting for tanθ in equation (7)
yields:
μs = 0.75
(b) Substitute numerical values in equation (6) and evaluate T:
(
)
1 kg
3.281 ft ⎞
⎛ lb
2
×
⎜ 22 ×
⎟ (4.0 m ) 9.81 m/s
ft 2.205 lb
m ⎠
T=⎝
= 1.1 kN
⎛ −1 ⎛ 3 ⎞ ⎞
2 sin ⎜⎜ tan ⎜ ⎟ ⎟⎟
⎝ 4 ⎠⎠
⎝
Remarks: 1.1 kN is approximately 240 lb.
59 ••
[SSM] Consider a rigid 2.5-m-long beam (Figure 12-51) that is
supported by a fixed 1.25-m-high post through its center and pivots on a frictionless
bearing at its center atop the vertical 1.25-m-high post. One end of the beam is
connected to the floor by a spring that has a force constant k = 1250 N/m. When the
beam is horizontal, the spring is vertical and unstressed. If an object is hung from
the opposite end of the beam, the beam settles into an equilibrium position where it
makes an angle of 17.5o with the horizontal. What is the mass of the object?
Picture the Problem Because the beam is in rotational equilibrium, we can apply
∑τ = 0 to it to determine the mass of the object suspended from its left end.
1
2
L
L
θ)
r
mg
r
Fby spring
1
0
1
2
n( 2
θ
x
L si
L
L − 12 L cos θ
α
L
1
2
1
2
ϕ
r
Fby bearing
θ
1
2
θ
α
δ
The pictorial representation directly
above shows the forces acting on the
beam when it is in static equilibrium.
The pictorial representation to the right
is an enlarged view of the right end of
the beam. We’ll use this diagram to
determine the length of the stretched
spring.
1
2
L
1
2
L sinθ
β
rstretched
1
2
ϕ
L
1238 Chapter 12
Apply
mg ( 12 L cos θ ) − Fby spring ( 12 L sin ϕ ) = 0
∑τ = 0 to the beam about an
or, because Fby spring = kΔ
axis through the bearing point to
obtain:
mg cos θ − kΔ
spring
spring
sin ϕ = 0
,
(1)
⎛ 1 L − 12 L cos θ ⎞
⎟⎟
1
⎝ 2 L + 2 L sin θ ⎠
⎛ 1 − cos θ ⎞
= tan −1 ⎜
⎟
⎝ 1 + sin θ ⎠
Use the right-hand diagram above to
relate the angles ϕ and θ :
ϕ = tan −1 ⎜⎜ 21
Substitute for ϕ in equation (1) to obtain:
mg cosθ − kΔ
spring
⎛
⎛ 1 − cosθ ⎞ ⎞
sin ⎜⎜ tan −1 ⎜
⎟ ⎟⎟ = 0
⎝ 1 + sin θ ⎠ ⎠
⎝
Solve for m to obtain:
kΔ
spring
m=
Δ
spring
Δ
is given by:
spring
=
⎛
⎛ 1 − cosθ ⎞ ⎞
sin ⎜⎜ tan −1 ⎜
⎟ ⎟⎟
⎝ 1 + sin θ ⎠ ⎠
⎝
(2)
g cosθ
stretched
−
unstretched
or, because unstretched = 12 L,
Δ spring = stretched − 12 L
To find the value of
stretched ,
refer to
2α + θ = π ⇒ α =
the right-hand diagram and note that:
Again, referring to the diagram,
relate β to α:
β =α +
Substituting for α yields:
β=
π
1
− θ
2 2
π
2
π
1
1
π
− θ + =π − θ
2 2
2
2
Apply the law of cosines to the triangle defined with bold sides:
2
stretched
= ( 12 L ) + (L sin 12 θ ) − 2( 12 L )(L sin 12 θ )cos(π − 12 θ )
2
Use the formula for the cosine of the
difference of two angles to obtain:
2
cos(π − 12 θ ) = − cos 12 θ
(3)
Static Equilibrium and Elasticity 1239
Substituting for cos(π − 12 θ ) yields:
2
stretched
= ( 12 L ) + (L sin 12 θ ) + 2( 12 L )(L sin 12 θ )cos 12 θ
2
2
= 14 L2 + L2 sin 2 12 θ + 12 L2 (2 sin 12 θ cos 12 θ )
Use the trigonometric identities sin θ = 2 sin 12 θ cos 12 θ and sin 2 12 θ =
1 − cos θ
to
2
obtain:
2
stretched
⎛ 1 − cos θ ⎞ 1 2
= 14 L2 + L2 ⎜
⎟ + 2 L (sin θ )
2
⎝
⎠
2
stretched
Simplifying yields:
or
stretched
Substituting for
Δ
spring
stretched
= 14 L2 [3 − 2(cos θ − sin θ )]
= 12 L 3 − 2(cos θ − sin θ )
in equation (3) yields:
(
)
= 12 L 3 − 2(cosθ − sin θ ) − 12 L = 12 L 3 − 2(cosθ − sin θ ) − 1
Substitute for Δ
spring
in equation (2) to obtain:
(
)
⎛
⎛ 1 − cosθ ⎞ ⎞
kL 3 − 2(cosθ − sin θ ) − 1 sin ⎜⎜ tan −1 ⎜
⎟ ⎟⎟
⎝ 1 + sin θ ⎠ ⎠
⎝
m=
g cosθ
1
2
Substitute numerical values and evaluate m:
1
2
(1250 N/m )(2.5 m )(
m=
)
⎛
⎛ 1 − cos17.5° ⎞ ⎞
3 − 2(cos17.5° − sin 17.5°) − 1 sin ⎜⎜ tan −1 ⎜
⎟ ⎟⎟
⎝ 1 + sin 17.5° ⎠ ⎠
⎝
(9.81 m/s 2 )cos17.5°
= 1.8 kg
60 ••
A rope and pulley system, called a block and tackle, is used to raise an
object of mass M (Figure 12-52) at constant speed. When the end of the rope moves
downward through a distance L, the height of the lower pulley is increased by h.
1240 Chapter 12
(a) What is the ratio L/h? (b) Assume that the mass of the block and tackle is
negligible and that the pulley bearings are frictionless. Show that FL = mgh by
applying the work–energy principle to the block–tackle object.
Picture the Problem We can determine the ratio of L to h by noting the number
of ropes supporting the load whose mass is M.
(a) Noting that three ropes support
the pulley to which the object whose
mass is M is fastened we can
conclude that:
L
= 3
h
(b) Apply the work-energy principle
to the block-tackle object to obtain:
Wext = ΔEsystem = ΔU block-tackle
or
FL = mgh
61 ••
A plate of mass M in the shape of an equilateral triangle is suspended
from one corner and a mass m is suspended from another of its corners. If the base
of the triangle makes an angle of 6.0º with the horizontal, what is the ratio m/M?
Picture the Problem The figure
shows the equilateral triangle without
the mass m, and then the same triangle
with the mass m and rotated through an
angle θ. Let the side length of the
triangle to be 2a. Then the center of
mass of the triangle is at a distance of
2a
from each vertex. As the triangle
3
rotates, its center of mass shifts
2a
θ, for θ << 1. Also, the vertex
by
3
to which m is attached moves toward
the plumb line by the distance d = 2aθ
cos30° = 3aθ (see the drawing).
Apply
∑τ
= 0 about an axis
through the point of suspension:
Solving for m/M yields:
(
)
mg a − 3aθ − Mg
2θ
m
=
M
3 1 − 3θ
(
)
2a
θ =0
3
Static Equilibrium and Elasticity 1241
Substitute numerical values and
evaluate m/M:
⎛ π rad ⎞
2(6.0°)⎜
⎟
m
180° ⎠
⎝
= 0.15
=
M
⎡
⎛ π rad ⎞⎤
3 ⎢1 − 3 (6.0°)⎜
⎟⎥
⎝ 180° ⎠⎦
⎣
62 ••
A standard six-sided pencil is placed on a pad of paper (Figure 12-53)
Find the minimum coefficient of static friction μs such that, if the pad is inclined,
the pencil rolls down the incline rather than sliding.
y
Picture the Problem If the hexagon is
to roll rather than slide, the incline’s
angle must be such that the center of
mass falls just beyond the support base.
From the geometry of the hexagon, the
critical angle is 30°. The free-body
diagram shows the forces acting on the
hexagonal pencil when it is on the verge
of sliding. We can use Newton’s 2nd law
to relate the coefficient of static friction
to the angle of the incline for which
rolling rather than sliding occurs.
Apply
∑ F = 0 to the pencil:
r
f s,max
r
Fn
θ
x
r
mg
∑F
x
= mg sin θ − f s,max = 0
(1)
and
∑ Fy = Fn − mg cosθ = 0
(2)
Substitute fs,max = µsFn in equation
(1):
mg sin θ − μs Fn = 0
(3)
Divide equation (3) by equation
(2) to obtain:
tan θ = μs
Thus, if the pencil is to roll rather
than slide when the pad is inclined:
μs ≥ tan 30° = 0.58
63 ••
An 8.0-kg box that has a uniform density and is twice as tall as it is wide
rests on the floor of a truck. What is the maximum coefficient of static friction
between the box and floor so that the box will slide toward the rear of the truck
rather than tip when the truck accelerates forward on a level road?
Picture the Problem The box and the forces acting on it are shown in the figure.
The force accelerating the box is the static friction force. When the box is about to
1242 Chapter 12
tip, Fn acts at its edge, as indicated in the drawing. We can use the definition of µs
and apply the condition for rotational equilibrium in an accelerated frame to relate fs
to the weight of the box and, hence, to the normal force.
w
2w
r
Fn
r
fs
Using its definition, express µs:
Apply
∑τ
= 0 about an axis through
the box’s center of mass:
Substitute for
fs
to obtain the
Fn
r
mg
μs ≥
fs
Fn
wf s − 12 wFn = 0 ⇒
fs 1
=
Fn 2
μs ≥ 0.50
condition for tipping:
Therefore, if the box is to slide:
μs < 0.50
64 ••
A balance scale has unequal arms. The scale is balanced with a 1.50-kg
block on the left pan and a 1.95 kg block on the right pan (Figure 12-54). If the
1.95-kg block is removed from the right pan and the 1.50-kg block is then moved to
the right pan, what mass on the left pan will balance the scale?
Picture the Problem Because the balance is in equilibrium, we can use the
condition for rotational equilibrium to relate the masses of the blocks to the lever
arms of the balance in the two configurations described in the problem statement.
Apply
∑τ
= 0 about an axis
through the fulcrum:
(1.50 kg )L1 − (1.95 kg )L2 = 0
Static Equilibrium and Elasticity 1243
Solving for L1 L2 yields:
Apply
∑τ
= 0 about an axis
L1
= 1.30
L2
ML1 − (1.50 kg )L2 = 0
through the fulcrum with 1.50 kg at
L2:
Solving for M yields:
Substitute for L1/L2 and evaluate M:
(1.50 kg )L2
M=
L1
=
1.50 kg
L1 L2
1.50 kg
= 1.15 kg
1.30
M=
65 ••
[SSM] A cube leans against a frictionless wall making an angle of θ
with the floor as shown in Figure 12-55. Find the minimum coefficient of static
friction μs between the cube and the floor that is needed to keep the cube from
slipping.
Picture the Problem Let the mass of
the cube be M. The figure shows the
location of the cube’s center of mass
and the forces acting on the cube. The
opposing couple is formed by the
friction force fs,max and the force exerted
by the wall. Because the cube is in
equilibrium, we can use the condition
for translational equilibrium to establish
that f s,max = FW and Fn = Mg and the
y
d
Mg
fs, max
∑ F = 0 to the cube:
θ
P
∑F
y
= Fn − Mg = 0 ⇒ Fn = Mg
and
∑ Fx = fs − FW = 0 ⇒ FW = fs
Noting that f s,max and FW form a
couple (their magnitudes are equal),
as do Fn and Mg, apply ∑τ = 0
about an axis though point P to
obtain:
FW
Fn
condition for rotational equilibrium to
relate the opposing couples.
Apply
a
f s,max a sin θ − Mgd = 0
a sin θ
x
1244 Chapter 12
Referring to the diagram to the right,
a
sin (45° + θ ) .
note that d =
2
a
2
r
Mg
r
Fn
45°
θ
d
Substitute for d and fs,max to obtain:
μs Mga sin θ − Mg
a
sin (45° + θ ) = 0
2
or
μs sin θ −
1
sin (45° + θ ) = 0
2
Solve for µs and simplify to obtain:
μs =
=
1
1
(sin 45° cosθ + cos 45° sin θ )
sin (45° + θ ) =
2 sin θ
2 sin θ
1
1
1
⎛ 1
⎞
cosθ +
sin θ ⎟ = (cot θ + 1)
⎜
2
2 sin θ ⎝ 2
2
⎠
66 ••
Figure 12-56 shows a 5.00-kg rod hinged to a vertical wall and
supported by a thin wire. The wire and rod each make angles of 45º with the
vertical. When a 10.0-kg block is suspended from the midpoint of the rod, the
tension T in the supporting wire is 52.0 N. If the wire will break when the tension
exceeds 75 N, what is the maximum distance from the hinge at which the block can
be suspended?
Picture the Problem Because the rod is in equilibrium, we can apply the condition
for rotational equilibrium to find the maximum distance from the hinge at which the
block can be suspended.
Apply
∑τ
= 0 about an axis through the hinge to obtain:
(1.00 m )(75 N ) − (0.50 m )(5.00 kg ) (9.81 m/s 2 )cos 45°
− d (10.0 kg ) (9.81 m/s 2 )cos 45° = 0
Solving for d yields:
d = 83 cm
Static Equilibrium and Elasticity 1245
67 ••
[SSM]
Figure 12-57 shows a 20.0-kg ladder leaning against a
frictionless wall and resting on a frictionless horizontal surface. To keep the ladder
from slipping, the bottom of the ladder is tied to the wall with a thin wire. When no
one is on the ladder, the tension in the wire is 29.4 N. (The wire will break if the
tension exceeds 200 N.) (a) If an 80.0-kg person climbs halfway up the ladder, what
force will be exerted by the ladder against the wall? (b) How far from the bottom
end of the ladder can an 80.0-kg person climb?
Picture the Problem Let m represent the
mass of the ladder and M the mass of the
person. The force diagram shows the
forces acting on the ladder for Part (b).
From the condition for translational
equilibrium, we can conclude that
T = Fby wall, a result we’ll need in Part
(b). Because the ladder is also in
rotational equilibrium, summing the
torques about the bottom of the ladder
will eliminate both Fn and T.
(a) Apply
∑τ
r
r
Fby wall
1
2
r
r
Fn
r
mg
r
Mg
0
θ
r
T
= 0 about an axis through the bottom of the ladder:
Fby wall ( sin θ ) − mg ( 12 cos θ ) − Mg ( 12 cos θ ) = 0
Solve for Fby wall and simplify to
obtain:
Refer to Figure 12-57 to determine θ :
Substitute numerical values and
evaluate Fby wall :
Fby wall =
(m + M )g cosθ = (m + M )g
2 sin θ
⎛ 5.0 m ⎞
⎟ = 73.3°
⎝ 1.5 m ⎠
θ = tan −1 ⎜
Fby wall =
(20 kg + 80 kg )(9.81 m/s2 )
2 tan 73.3°
= 0.15 kN
(b) Apply
obtain:
∑F
x
= 0 to the ladder to
2 tan θ
T − Fby wall = 0 ⇒ T = Fby wall
1246 Chapter 12
Apply
∑τ
= 0 about an axis through the bottom of the ladder subject to the
condition that Fby wall = Tmax :
Tmax ( sin θ ) − mg ( 12 cosθ ) − Mg (L cosθ ) = 0
where L is the maximum distance along the ladder that the person can climb without
exceeding the maximum tension in the wire.
Solving for L and simplifying yields:
L=
Tmax sin θ − 12 mg cosθ
Mg cosθ
Substitute numerical values and evaluate L:
L=
(200 N )(5.0 m ) − 12 (20 kg ) (9.81 m/s 2 )(1.5 m ) =
(80 kg ) (9.81 m/s 2 )cos 73.3°
3.8 m
68 ••
A 360-kg object is supported on a wire attached to a 15-m-long steel bar
that is pivoted at a vertical wall and supported by a cable as shown in Figure 12-58.
The mass of the bar is 85 kg. With the cable attached to the bar 5.0 m from the
lower end as shown, find the tension in the cable and the force exerted by the wall
on the steel bar.
Picture the Problem Let m represent
the mass of the bar, M the mass of the
suspended object, Fv the vertical
component of the force the wall exerts
on the bar, Fh the horizontal component
of the force the wall exerts on the bar,
and T the tension in the cable. The force
diagram shows these forces and their
points of application on the bar. Because
the bar is in equilibrium, we can apply
the conditions for translational and
rotational equilibrium to relate the
various forces and distances.
Apply
∑τ
= 0 to the bar about an
r3
r
T
y
r
Fv
0
T
r1
θ
r
Fh
1
r2
θ
r
Mg
r
mg
x
cosθ − Mg
− mg
2
(m
+M
axis through the hinge:
Solving for T yields:
T=
2
3
1
)g cosθ
3
cosθ = 0
Static Equilibrium and Elasticity 1247
Substitute numerical values and evaluate T:
T=
((85 kg )(7.5 m ) + (360 kg )(15 m )) (9.81 m/s 2 )cos 30° = 10.3 kN =
5.0 m
The magnitude and direction of the
force exerted by the wall are given
by:
Fby wall = Fv2 + Fh2
(1)
and
⎛ Fv ⎞
⎟⎟
⎝ Fh ⎠
φ = tan −1 ⎜⎜
Apply
10 kN
∑F
∑ F = 0 to the bar to obtain:
y
(2)
= Fv + T cos 30° − mg − Mg = 0
and
∑ Fx = Fh − T sin 30° = 0
Solving the y equation for Fv yields:
Fv = (m + M )g − T cos 30°
Solve the x equation for Fh to obtain:
Fh = T sin 30°
Substituting for Fv and Fh in equation (1) yields:
Fby wall =
((m + M )g − T cos 30°)2 + (T sin 30°)2
Substitute numerical values and evaluate Fby wall:
Fby wall =
((85 kg + 360 kg )(9.81 m/s )− (10.3 kN )cos 30°) + ((10.3 kN )sin 30°)
2
2
= 6.9 kN
Substituting for Fv and Fh in equation
(2) yields:
⎛ (m + M )g − T cos 30° ⎞
⎟
T sin 30°
⎝
⎠
φ = tan −1 ⎜
Substitute numerical values and evaluate φ:
(
)
⎛ (85 kg + 360 kg ) 9.81 m/s 2 − (10.3 kN )cos 30° ⎞
⎟⎟ = −41.49°
(10.3 kN )sin 30°
⎝
⎠
φ = tan −1 ⎜⎜
= − 41° That is, 41° below the horizontal.
2
1248 Chapter 12
69 ••
Repeat Problem 63 if the truck accelerates up a hill that makes an angle
of 9.0º with the horizontal.
y
Picture the Problem The box and the
forces acting on it are shown in the
figure. When the box is about to tip, Fn
acts at its edge, as indicated in the
drawing. We can use the definition of µs
and apply the condition for rotational
equilibrium in an accelerated frame to
relate fs to the weight of the box and,
hence, to the normal force.
w
2w
+
r
Fn
r
mg
r
fs
Using its definition, express µs:
Apply
∑τ
= 0 about an axis
through the box’s center of mass:
Substitute for
fs
to obtain the
Fn
μs ≥
θ
x
θ
fs
Fn
wf s − 12 wFn = 0 ⇒
fs 1
=
Fn 2
μs ≥ 0.50
condition for tipping:
Therefore, if the box is to slide:
μs < 0.50 , as in Problem 63.
Remarks: The difference between problems 63 and 69 is that in 63 the
maximum acceleration before slipping is 0.5g, whereas in 69 it is
(0.5 cos9.0°− sin9.0°)g = 0.337g.
70 ••
A thin uniform rod 60 cm long is balanced 20 cm from one end when an
object whose mass is (2m + 2.0 grams) is at the end nearest the pivot and an object
of mass m is at the opposite end (Figure 12-59a). Balance is again achieved if the
object whose mass is (2m + 2.0 grams) is replaced by the object of mass m and no
object is placed at the other end (Figure 12-59b). Determine the mass of the rod.
Picture the Problem Let the mass of the rod be represented by M. Because the rod
is in equilibrium, we can apply the condition for rotational equilibrium to relate the
masses of the objects placed on the rod to its mass.
Static Equilibrium and Elasticity 1249
the pivot for the initial condition:
(20 cm )(2m + 2.0 g ) − (40 cm ) m
− (10 cm ) M = 0
Simplifying yields:
2 (2m + 2.0 g ) − 4 m − M = 0
Solve for M to obtain:
M = 4.0 g
Apply
∑τ
= 0 about an axis through
71 ••• [SSM] There are a large number of identical uniform bricks, each of
length L. If they are stacked one on top of another lengthwise (see Figure 12-60),
the maximum offset that will allow the top brick to rest on the bottom brick is L/2.
(a) Show that if this two-brick stack is placed on top of a third brick, the maximum
offset of the second brick on the third brick is L/4. (b) Show that, in general, if you
build a stack of N bricks, the maximum overhang of the (n − 1)th brick (counting
down from the top) on the nth brick is L/2n. (c) Write a spreadsheet program to
calculate total offset (the sum of the individual offsets) for a stack of N bricks, and
calculate this for L = 20 cm and N = 5, 10, and 100. (d) Does the sum of the
individual offsets approach a finite limit as N → ∞? If so, what is that limit?
Picture the Problem Let the weight of each uniform brick be w. The downward
force of all the bricks above the nth brick must act at its corner, because the
upward reaction force points through the center of mass of all the bricks above the
nth one. Because there is no vertical acceleration, the upward force exerted by the
(n + 1)th brick on the nth brick must equal the total weight of the bricks above it.
Thus this force is just nw. Note that it is convenient to develop the general
relationship of Part (b) initially and then extract the answer for Part (a) from this
general result.
downward force from n − 1
blocks above; their center
of mass is directly above
this edge
1
in
P
w of
block n
n+ 1
N
dn
upward force
from corner of
block n + 1 = nw
1250 Chapter 12
w( 12 L ) − nwd n = 0
(a) and (b) Noting that the line of
action of the downward forces
exerted by the blocks above the nth
block passes through the point P,
resulting in a lever arm of zero, apply
∑τ P = 0 to the nth brick to obtain:
where dn is the overhang of the nth
brick beyond the edge of the (n + 1)th
brick.
Solving for dn yields:
For block number 2:
dn =
L
where n = 1,2,3,…
2n
d2 =
L
4
(c) A spreadsheet program to calculate the sum of the offsets as a function of n is
shown below. The formulas used to calculate the quantities in the columns are as
follows:
Cell Formula/Content Algebraic Form
B5
B4+1
n+1
C5 C4+$B$1/(2*B5)
L
dn +
2n
A
1
2
3
4
5
6
7
8
9
10
11
12
13
98
99
100
101
102
103
B
L= 0.20
C
m
n
1
2
3
4
5
6
7
8
9
10
offset
0.100
0.150
0.183
0.208
0.228
0.245
0.259
0.272
0.283
0.293
95
96
97
98
99
100
0.514
0.515
0.516
0.517
0.518
0.519
D
Static Equilibrium and Elasticity 1251
From the table we see that d5 = 15 cm, d 10 = 26 cm, and
d 100 = 0.52 cm.
N
(d) The sum of the individual offsets
S is given by:
S = ∑ dn =
n =1
L N 1
∑
2 n =1 n
Because this series is a harmonic series, S approaches infinity as the number of
blocks N grows without bound. The following graph, plotted using a spreadsheet
program, suggests that S has no limit.
Offset as a function of n for L = 20 cm
0.6
0.5
Offset, m
0.4
0.3
0.2
0.1
0.0
0
20
40
60
80
100
n
72 ••
A uniform sphere of radius R and mass M is held at rest on an inclined
plane of angle θ by a horizontal string, as shown in Figure 12-61. Let R = 20 cm,
M = 3.0 kg, and θ = 30º. (a) Find the tension in the string. (b) What is the normal
force exerted on the sphere by the inclined plane? (c) What is the frictional force
acting on the sphere?
1252 Chapter 12
Picture the Problem The four forces
acting on the sphere: its weight, mg; the
normal force of the plane, Fn; the
frictional force, f, acting parallel to the
plane; and the tension in the string, T,
are shown in the figure. Because the
sphere is in equilibrium, we can apply
the conditions for translational and
rotational equilibrium to find f, Fn, and
T.
(a) Apply
∑τ
= 0 about an axis
θ
T
y
θ
x
Mg
f
Fn
fR − TR = 0 ⇒ T = f
through the center of the sphere:
Apply
∑F
x
= 0 to the sphere:
f + T cos θ − Mg sin θ = 0
Substituting for f and solving for
T yields:
T=
Substitute numerical values and
evaluate T:
T=
Mg sin θ
1 + cosθ
(3.0 kg ) (9.81 m/s 2 )sin 30° = 7.89 N
1 + cos30°
= 7 .9 N
(b) Apply
∑F
y
= 0 to the sphere:
Solve for Fn:
Substitute numerical values and
evaluate Fn:
Fn − T sin θ − Mg cos θ = 0
Fn = T sin θ + Mg cos θ
Fn = (7.89 N )sin30°
(
)
+ (3.0 kg ) 9.81 m/s 2 cos 30°
= 29 N
(c) In Part (a) we showed that f = T:
f = 7 .9 N
73 ••• The legs of a tripod make equal angles of 90º with each other at the
apex, where they join together. A 100-kg block hangs from the apex. What are the
compressional forces in the three legs?
Static Equilibrium and Elasticity 1253
Picture the Problem Let L be the
length of each leg of the tripod.
Applying the Pythagorean theorem
leads us to conclude that the distance a
shown in the figure is 3 2 L and the
distance b, the distance to the centroid
of the triangle ABC is 23 3 2 L , and the
distance c is L
3 . These results allow
us to conclude that cos θ = L 3 .
Because the tripod is in equilibrium, we
can apply the condition for translational
equilibrium to find the compressional
forces in each leg.
Letting FC represent the
compressional force in a leg of the
tripod, apply ∑ F = 0 to the apex of
3FC cos θ − mg = 0 ⇒ FC =
mg
3 cos θ
the tripod:
Substitute for cosθ and simplify to
obtain:
FC =
Substitute numerical values and
evaluate FC:
FC =
3
mg
=
mg
3
⎛ 1 ⎞
3⎜
⎟
⎝ 3⎠
(
)
3
(100 kg ) 9.81m/s2 = 566 N
3
74 ••• Figure 12-63 shows a 20-cm-long uniform beam resting on a cylinder
that has a radius of 4.0-cm. The mass of the beam is 5.0 kg and that of the cylinder
is 8.0 kg. The coefficient of static friction between beam and cylinder is zero,
whereas the coefficients of static friction between the cylinder and the floor, and
between the beam and the floor, are not zero. Are there any values for these
coefficients of static friction such that the system is in static equilibrium? If so,
what are these values? If not, explain why none exist.
Picture the Problem The forces that act on the beam are its weight, mg; the force
of the cylinder, Fc, acting along the radius of the cylinder; the normal force of the
ground, Fn; and the friction force fs = µsFn. The forces acting on the cylinder are its
weight, Mg; the force of the beam on the cylinder, Fcb = Fc in magnitude, acting
radially inward; the normal force of the ground on the cylinder, Fnc; and the force of
friction, fsc = µscFnc. Choose the coordinate system shown in the figure and apply
the conditions for rotational and translational equilibrium.
1254 Chapter 12
y
θ
Fc
ο
90 −θ
θ
x
Fcb
Fn
mg
Mg
θ
fs
fsc
Fnc
Express μs,beam−floor in terms of fs and
Fn:
Express μs,cylinder−floor in terms of fsc
and Fnc:
Apply
∑τ
= 0 about an axis through
μs, beam − floor =
fs
Fn
μs,cylinder − floor =
(1)
f sc
Fnc
(2)
[(10 cm)cosθ ]mg − (15 cm)Fc = 0
the right end of the beam:
Solve for Fc to obtain:
Substitute numerical values and
evaluate Fc:
Fc =
Fc
[(10 cm)cosθ ]mg
15 cm
10 cos 30°](5.0 kg ) (9.81 m/s 2 )
[
=
15
= 28.3N
Apply
∑F
y
= 0 to the beam:
Solving for Fn yields:
Substitute numerical values and
evaluate Fn:
Apply
∑F
x
= 0 to the beam:
Solve for fs to obtain:
Fn + Fc cosθ − mg = 0
Fn = mg − Fc cosθ
Fn = (5.0 kg ) (9.81 m/s 2 ) − (28.3 N ) cos30°
= 24.5 N
− fs + Fc cos(90° − θ ) = 0
f s = Fc cos(90° − θ )
Static Equilibrium and Elasticity 1255
Substitute numerical values and
evaluate fs:
Fcb is the reaction force to Fc :
Apply
∑F
y
= 0 to the cylinder:
Solve for Fnc to obtain:
Substitute numerical values and
evaluate Fnc:
Apply
∑F
x
= 0 to the cylinder:
Solve for and evaluate fsc:
f s = Fc cos(90° − θ ) = (28.3 N ) cos60°
= 14.2 N
Fcb = Fc = 28.3 N radially inward.
Fnc − Fcb cos θ − Mg = 0
Fnc = Fcb cosθ + Mg
Fnc = (28.3 N ) cos30° + (8.0 kg ) (9.81 m/s 2 )
= 103 N
fsc − Fcb cos(90° − θ ) = 0
f sc = Fcb cos(90° − θ ) = (28.3 N )cos60°
= 14.2 N
Substitute numerical values in
equations (1) and (2) and evaluate
µs,beam-floor and µs,cylinder-floor:
μs,beam−floor =
14.2 N
= 0.58
24.5 N
and
μs,cylinder−floor =
14.2 N
= 0.14
103 N
75 ••• [SSM] Two solid smooth (frictionless) spheres of radius r are placed
inside a cylinder of radius R, as in Figure 12-62. The mass of each sphere is m. Find
the force exerted by the bottom of the cylinder on the bottom sphere, the force
exerted by the wall of the cylinder on each sphere, and the force exerted by one
sphere on the other. All forces should be expressed in terms of m, R, and r.
1256 Chapter 12
Picture the Problem The geometry of
the system is shown in the drawing.
Let upward be the positive y direction
and to the right be the positive x
direction. Let the angle between the
vertical center line and the line joining
the two centers be θ. Then
R−r
R−r
sin θ =
and tan θ =
.
r
R(2r − R )
FW
θ
mg
θ F
The force exerted by the bottom of the
cylinder is just 2mg. Let F be the force
that the top sphere exerts on the lower
sphere. Because the spheres are in
equilibrium, we can apply the
condition for translational equilibrium.
Apply
∑F
y
= 0 to the spheres:
r
r
F' W
R–r
mg
R
Fn
Fn − mg − mg = 0 ⇒ Fn = 2mg
Because the cylinder wall is smooth,
Fcosθ = mg, and:
F=
Express the x component of F:
Fx = F sin θ = mg tan θ
Express the force that the wall of the
cylinder exerts:
FW = mg
mg
r
= mg
cosθ
R(2r − R )
R−r
R(2r − R )
Remarks: Note that as r approaches R/2, Fw→∞.
76 ••• A solid cube of side length a balanced atop a cylinder of diameter d is in
unstable equilibrium if d << a (Figure 12-64) and is in stable equilibrium if d >> a.
Determine the minimum value of the ratio d/a for which the cube is in stable
equilibrium.
Picture the Problem Consider a small rotational displacement, δθ of the cube
from equilibrium. This shifts the point of contact between cube and cylinder
by Rδθ, where R = d/2. As a result of that motion, the cube itself is rotated through
the same angle δθ, and so its center is shifted in the same direction by the amount
(a/2) δθ, neglecting higher order terms in δθ .
Static Equilibrium and Elasticity 1257
+
a
+
δθ
R
d
If the displacement of the cube’s center of mass is less than that of the point of
contact, the torque about the point of contact is a restoring torque, and the cube will
return to its equilibrium position. If, on the other hand, (a/2) δθ > (d/2) δθ, then the
torque about the point of contact due to mg is in the direction of δθ, and will cause
the displacement from equilibrium to increase. We see that the minimum value of
d/a for stable equilibrium is
d/a = 1.
1258 Chapter 12